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1970 | 1970_B1 | Evaluate $\lim_{n \to \infty} \frac{1}{n^4} \prod_{i=1}^{2n} (n^2 + i^2)^{1/n}.$ | Let \[ a_n = \frac{1}{n^4} \prod_{i=1}^{2n} (n^2 + i^2)^{1/n}. \] Then \[ \log a_n = \frac{1}{n} \sum_{i=1}^{2n} \log \left( 1 + \frac{i^2}{n^2} \right), \] and \[ \lim_{n \to \infty} \log a_n = \int_0^2 \log(1 + x^2)dx = \boxed{2 \log 5 - 4 + 2 \arctan 2}. \] | numerical | putnam | Calculus Analysis | Evaluate $\lim_{n \to \infty} \frac{1}{n^4} \prod_{i=1}^{2n} (n^2 + i^2)^{1/n}.$ | Let \[ a_n = \frac{1}{n^4} \prod_{i=1}^{2n} (n^2 + i^2)^{1/n}. \] Then \[ \log a_n = \frac{1}{n} \sum_{i=1}^{2n} \log \left( 1 + \frac{i^2}{n^2} \right), \] and \[ \lim_{n \to \infty} \log a_n = \int_0^2 \log(1 + x^2)dx = \boxed{2 \log 5 - 4 + 2 \arctan 2}. \] | 0 |
1970 | 1970_B2 | The time-varying temperature of a certain body is given by a polynomial in the time of degree at most three. Given that the average temperature of the body between $9 \text{A.M.}$ and $3 \text{P.M.}$ can always be found by taking the average of the temperatures at two fixed times, which are independent of which polynomial occurs find these times to the nearest minute and return the sums of both hour and minute values when expressed in 24-hour format. | Let $P(t) = at^3 + bt^2 + ct + d$. The equation
\[ \frac{1}{2T} \int_{-T}^T P(t) \, dt = \frac{1}{2} \{P(t_1) + P(t_2)\} \]
is satisfied for all values of $a, b, c, \text{ and } d$ if and only if $t_2 = -t_1 = \pm T/\sqrt{3}$. If $T = 3 \text{ hrs}, T/\sqrt{3} \approx 1 \text{ hr}, 43.92 \text{ min}$. Therefore, in the case considered, the critical times are $1 \text{ hour } 44 \text{ minutes}$ each side of noon. With this information the final answer becomes \boxed{83}. | numerical | putnam (modified boxing) | Analysis Calculus | The time-varying temperature of a certain body is given by a polynomial in the time of degree at most three. Show that the average temperature of the body between $9 \text{A.M.}$ and $3 \text{P.M.}$ can always be found by taking the average of the temperatures at two fixed times, which are independent of which polynomial occurs. Also, show that these two times are $10:16 \text{A.M.}$ and $1:44 \text{P.M.}$ to the nearest minute. | Let $P(t) = at^3 + bt^2 + ct + d$. The equation
\[ \frac{1}{2T} \int_{-T}^T P(t) \, dt = \frac{1}{2} \{P(t_1) + P(t_2)\} \]
is satisfied for all values of $a, b, c, \text{ and } d$ if and only if $t_2 = -t_1 = \pm T/\sqrt{3}$. If $T = 3 \text{ hrs}, T/\sqrt{3} \approx 1 \text{ hr}, 43.92 \text{ min}$. Therefore, in the case considered, the critical times are $1 \text{ hour } 44 \text{ minutes}$ each side of noon. | 0 |
1971 | 1971_A3 | The three vertices of a triangle of sides $a, b, \text{and } c$ are lattice points and lie on a circle of radius $R$. Find the minimum value of $abc. (Lattice points are points in the Euclidean plane with integral coordinates.) | For a triangle with sides $a, b, c$, area $= A$ and circumradius $= R$ we have $abc = 4RA$. But if the vertices are lattice points the determinant formula (or Pick’s Theorem or direct calculation) for the area shows that $2A$ is an integer. Hence $2A \geq 1$, so that $abc \geq \boxed{2R}$. To obtain the formula $abc = 4RA$ note that if $\alpha$ is the angle opposite side $a$, then side $a$ subtends an angle $2\alpha$ at the center and $a = 2R \sin \alpha$, $A = \frac{1}{2} bc \sin \alpha$. | algebraic | putnam (modified boxing) | Geometry Number Theory | The three vertices of a triangle of sides $a, b, \text{and } c$ are lattice points and lie on a circle of radius $R$. Show that $abc \geq 2R$. (Lattice points are points in the Euclidean plane with integral coordinates.) | For a triangle with sides $a, b, c$, area $= A$ and circumradius $= R$ we have $abc = 4RA$. But if the vertices are lattice points the determinant formula (or Pick’s Theorem or direct calculation) for the area shows that $2A$ is an integer. Hence $2A \geq 1$, so that $abc \geq 2R$. To obtain the formula $abc = 4RA$ note that if $\alpha$ is the angle opposite side $a$, then side $a$ subtends an angle $2\alpha$ at the center and $a = 2R \sin \alpha$, $A = \frac{1}{2} bc \sin \alpha$. | 0 |
1971 | 1971_A4 | Show that for $0 < \varepsilon < 1$ the expression $(x+y)^n (x^2 - (2-\varepsilon)x + y^2)$ is a polynomial with positive coefficients for $n$ sufficiently large and integral. For $\varepsilon = .002$ find the smallest admissible value of $n$. | In the expansion of $(x+y)^n(x^2-(2-\varepsilon)xy+y^2)$ the coefficient of $x^k y^{n+1-k}$ is
\[
\binom{n}{k-1}-(2-\varepsilon)\binom{n}{k}+\binom{n}{k+1}.
\]
\[= \binom{n}{k}\left[\frac{k}{n-k+1}+\frac{n-k}{k+1}-(2-\varepsilon)\right].\]
Now for fixed $n$ consider the expression
\[
\phi(k) = \frac{k}{n-k+1} + \frac{n-k}{k+1} - (2-\varepsilon).
\]
If $k$ is taken to be a continuous positive variable,
\[
\phi'(k) = \frac{(n+1)((k+1)^2 - (n-k+1)^2)}{(n-k+1)^2(k+1)^2}.
\]
Hence $\phi'(k) = 0$ at $k = n/2$ and it follows easily that $\phi(k)$ is minimum at $k = n/2$.
We needn't consider end point minima since it easily follows that for $n > 2$ the polynomial has its first two and last two coefficients positive. We may also note that if the two mid-terms in the expansion are non-positive for a given odd value of $n$ then for the next larger value of $n$ the mid-term remains non-positive. Hence if the mid-coefficients become positive, the first value of $n$ for which this occurs is odd. Now if
\[
n \text{ is odd and } k = \frac{1}{2}(n+1), \phi(k) = \frac{n-1}{n+3} - 1 + \varepsilon, \text{ and } \phi(k) > 0 \text{ for } n > \frac{4}{\varepsilon} - 3.
\]
If $\varepsilon = .002, n > 1997 \text{ and } n \text{ is odd.}$ Hence the minimum $n$ for which all terms are positive is $\boxed{1999}$. | numerical | putnam | Algebra | Show that for $0 < \varepsilon < 1$ the expression $(x+y)^n (x^2 - (2-\varepsilon)x + y^2)$ is a polynomial with positive coefficients for $n$ sufficiently large and integral. For $\varepsilon = .002$ find the smallest admissible value of $n$. | In the expansion of $(x+y)^n(x^2-(2-\varepsilon)xy+y^2)$ the coefficient of $x^k y^{n+1-k}$ is
\[
\binom{n}{k-1}-(2-\varepsilon)\binom{n}{k}+\binom{n}{k+1}.
\]
\[= \binom{n}{k}\left[\frac{k}{n-k+1}+\frac{n-k}{k+1}-(2-\varepsilon)\right].\]
Now for fixed $n$ consider the expression
\[
\phi(k) = \frac{k}{n-k+1} + \frac{n-k}{k+1} - (2-\varepsilon).
\]
If $k$ is taken to be a continuous positive variable,
\[
\phi'(k) = \frac{(n+1)((k+1)^2 - (n-k+1)^2)}{(n-k+1)^2(k+1)^2}.
\]
Hence $\phi'(k) = 0$ at $k = n/2$ and it follows easily that $\phi(k)$ is minimum at $k = n/2$.
We needn't consider end point minima since it easily follows that for $n > 2$ the polynomial has its first two and last two coefficients positive. We may also note that if the two mid-terms in the expansion are non-positive for a given odd value of $n$ then for the next larger value of $n$ the mid-term remains non-positive. Hence if the mid-coefficients become positive, the first value of $n$ for which this occurs is odd. Now if
\[
n \text{ is odd and } k = \frac{1}{2}(n+1), \phi(k) = \frac{n-1}{n+3} - 1 + \varepsilon, \text{ and } \phi(k) > 0 \text{ for } n > \frac{4}{\varepsilon} - 3.
\]
If $\varepsilon = .002, n > 1997 \text{ and } n \text{ is odd.}$ Hence the minimum $n$ for which all terms are positive is $\boxed{1999}$. | 0 |
1971 | 1971_A5 | A game of solitaire is played as follows. After each play, according to the outcome, the player receives either $a$ or $b$ points ($a$ and $b$ are positive integers with $a$ greater than $b$), and his score accumulates from play to play. It has been noticed that there are thirty-five non-attainable scores and that one of these is 58. Find $a+b$. | The attainable scores are those non-negative integers expressible in the form $ax + by$ with $x$ and $y$ non-negative integers. If $a$ and $b$ are not relatively prime there are infinitely many non-attainable scores. Hence $(a, b) = 1$. It will be shown that the number of non-attainable scores is $\frac{1}{2}(a - 1)(b - 1)$.
If $m$ is an attainable score, the line $ax + by = m$ passes through at least one lattice point in the closed first quadrant. Because $a$ and $b$ are relatively prime, the lattice points on a line $ax + by = m$ are at a horizontal distance of $b$. The first-quadrant segment of $ax + by = m$ has a horizontal projection of $m/a$ and thus every score $m \geq ab$ is attainable. Every non-attainable score must satisfy $0 \leq m < ab$.
If $0 \leq m < ab$, the first-quadrant segment of the line $ax + by = m$ has a horizontal projection less than $b$, and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points $(x, y)$ with $0 \leq ax + by < ab$ in the first quadrant and attainable scores with $0 \leq m < ab$. The closed rectangle $0 \leq x \leq b$, $0 \leq y \leq a$ contains $(a + 1)(b + 1)$ lattice points, so the number of lattice points in the first quadrant with $0 \leq ax + by < ab$ is $ab - \frac{1}{2}(a + 1)(b + 1) + 1 = \frac{1}{2}(a - 1)(b - 1)$.
In our given example $70 = (a - 1)(b - 1) = 1(70) = 2(35) = 5(14) = 7(10)$. The conditions $a > b$, $(a, b) = 1$ yield two possibilities $a = 71, b = 2$ and $a = 11, b = 8$. Since 58 = 71(0) + 2(29), the first of these alternatives is eliminated. The line $11x + 8y = 58$ passes through $(6, -1)$ and $(-2, 10)$ and thus does not pass through a lattice point in the first quadrant. The unique solution is $a = 11$, $b = 8$ making the final answer \boxed{19}. | numerical | putnam (modified boxing) | Number Theory | A game of solitaire is played as follows. After each play, according to the outcome, the player receives either $a$ or $b$ points ($a$ and $b$ are positive integers with $a$ greater than $b$), and his score accumulates from play to play. It has been noticed that there are thirty-five non-attainable scores and that one of these is 58. Find $a$ and $b$. | The attainable scores are those non-negative integers expressible in the form $ax + by$ with $x$ and $y$ non-negative integers. If $a$ and $b$ are not relatively prime there are infinitely many non-attainable scores. Hence $(a, b) = 1$. It will be shown that the number of non-attainable scores is $\frac{1}{2}(a - 1)(b - 1)$.
If $m$ is an attainable score, the line $ax + by = m$ passes through at least one lattice point in the closed first quadrant. Because $a$ and $b$ are relatively prime, the lattice points on a line $ax + by = m$ are at a horizontal distance of $b$. The first-quadrant segment of $ax + by = m$ has a horizontal projection of $m/a$ and thus every score $m \geq ab$ is attainable. Every non-attainable score must satisfy $0 \leq m < ab$.
If $0 \leq m < ab$, the first-quadrant segment of the line $ax + by = m$ has a horizontal projection less than $b$, and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points $(x, y)$ with $0 \leq ax + by < ab$ in the first quadrant and attainable scores with $0 \leq m < ab$. The closed rectangle $0 \leq x \leq b$, $0 \leq y \leq a$ contains $(a + 1)(b + 1)$ lattice points, so the number of lattice points in the first quadrant with $0 \leq ax + by < ab$ is $ab - \frac{1}{2}(a + 1)(b + 1) + 1 = \frac{1}{2}(a - 1)(b - 1)$.
In our given example $70 = (a - 1)(b - 1) = 1(70) = 2(35) = 5(14) = 7(10)$. The conditions $a > b$, $(a, b) = 1$ yield two possibilities $a = 71, b = 2$ and $a = 11, b = 8$. Since 58 = 71(0) + 2(29), the first of these alternatives is eliminated. The line $11x + 8y = 58$ passes through $(6, -1)$ and $(-2, 10)$ and thus does not pass through a lattice point in the first quadrant. The unique solution is $a = 11$, $b = 8$. | 0 |
1971 | 1971_B2 | Let $F(x)$ be a real valued function defined for all real $x$ except for $x=0$ and $x=1$ and satisfying the functional equation $F(x) + F\left(\frac{x-1}{x}\right) = 1 + x$. Find all functions $F(x)$ satisfying these conditions. | In the given functional equation \[(1)\ F(x) + F\left(\frac{x-1}{x}\right) = 1 + x\] we substitute $\frac{x-1}{x}$ for $x$, obtaining \[(2)\ F\left(\frac{x-1}{x}\right) + F\left(\frac{-1}{x-1}\right) = \frac{2x-1}{x}.\] Also in (1), we substitute $\frac{-1}{x-1}$ for $x$ and obtain \[(3)\ F\left(\frac{-1}{x-1}\right) + F(x) = \frac{x-2}{x-1}.\] Adding (1) and (3) and subtracting (2) gives \[2F(x) = 1 + x + \frac{x-2}{x-1} - \frac{2x-1}{x} = \frac{x^3 - x^2 - 1}{x(x-1)}.\] \[F(x) = \boxed{\frac{x^3 - x^2 - 1}{2x(x-1)}}.\] That $F(x)$, defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem. | algebraic | putnam | Algebra Analysis | Let $F(x)$ be a real valued function defined for all real $x$ except for $x=0$ and $x=1$ and satisfying the functional equation $F(x) + F\left(\frac{x-1}{x}\right) = 1 + x$. Find all functions $F(x)$ satisfying these conditions. | In the given functional equation \[(1)\ F(x) + F\left(\frac{x-1}{x}\right) = 1 + x\] we substitute $\frac{x-1}{x}$ for $x$, obtaining \[(2)\ F\left(\frac{x-1}{x}\right) + F\left(\frac{-1}{x-1}\right) = \frac{2x-1}{x}.\] Also in (1), we substitute $\frac{-1}{x-1}$ for $x$ and obtain \[(3)\ F\left(\frac{-1}{x-1}\right) + F(x) = \frac{x-2}{x-1}.\] Adding (1) and (3) and subtracting (2) gives \[2F(x) = 1 + x + \frac{x-2}{x-1} - \frac{2x-1}{x} = \frac{x^3 - x^2 - 1}{x(x-1)}.\] \[F(x) = \boxed{\frac{x^3 - x^2 - 1}{2x(x-1)}}.\] That $F(x)$, defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem. | 0 |
1971 | 1971_B4 | A "spherical ellipse" with foci $A, B$ on a given sphere is defined as the set of all points $P$ on the sphere such that $\widehat{PA} + \widehat{PB} = \text{constant}$. Here $\widehat{PA}$ denotes the shortest distance on the sphere between $P$ and $A$. Determine the entire class of real spherical ellipses which are circles by finding the \text{constant} value. | We take the radius of the sphere as unity and denote the constant sum $\widehat{PA} + \widehat{PB}$ by $2a$. To avoid trivial and degenerate cases we assume that $0 < \widehat{AB} < \pi$ and that $\widehat{AB} < 2a < 2\pi - \widehat{AB}$.
The case $2a > \pi$ can be reduced to the case $2a < \pi$. For, if $A'$ and $B'$ are the points diametrically opposite to $A$ and $B$ then $\widehat{PA} + \widehat{PB} = 2a$ if and only if $\widehat{PA}' + \widehat{PB}' = 2\pi - 2a$; that is, the spherical ellipses $\widehat{PA} + \widehat{PB} = 2a$ and $\widehat{PA}' + \widehat{PB}' = 2\pi - 2a$ are identical. Since $\min(2a, 2\pi - 2a) \leq \pi$, we may assume without loss of generality that $2a \leq \pi$.
Let $A$ and $B$ lie on the equator. There are two points $V_1$ and $V_2$ (the "vertices") on the equator which lie on the spherical ellipse. Obviously, $V_1V_2 = 2a$. The "center" of the spherical ellipse (common midpoint of the arcs $\widehat{AB}$ and $\widehat{V_1V_2}$) will be denoted by $C$.
We first treat the case $2a < \pi$ and show that in this case the spherical ellipse cannot be a circle. Assume it were a circle; call it $\Gamma$ (see Figure 1). $\Gamma$ would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. $\Gamma$ would also have to pass through the vertices. Therefore its spherical diameter would be $V_1V_2 = 2a$ and its spherical radius would be equal to $a$. The spherical center of $\Gamma$, the center of the ellipse, let $M$ be one of the two points on $\Gamma$ which lie half-way between the two vertices. Then, since $M$ is supposed to be a point on the spherical ellipse, $2a = \widehat{MA} + \widehat{MB} > \widehat{MC} = 2a$ (note that $\widehat{MC}$ is a right spherical triangle with the right angle at $C$ and with side $\widehat{MC} = a < 2a$). Contradiction shows that the only possible spherical ellipses which are circles must occur when $2a = \pi$.
In case $2a = \pi$, $V_1$ and $V_2$ are diametrically opposite points on the equator. We shall show that the great circle $\Gamma$ through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse $\widehat{PA} + \widehat{PB} = \pi$. To see this, let $B^*$ be the reflection of $B$ about the plane of $\Gamma$. $B^*$ is on the equator diametrically opposite to $A$ (see Fig. 2). Let $P$ be an arbitrary point on the sphere, and draw the great circle through $A$, $P$ and $B^*$. Then $\widehat{PA} + \widehat{PB} = \pi$ if and only if $\widehat{PB} = \widehat{PB^*}$; that is, if and only if $P$ is on $\Gamma$. This shows that $\Gamma$ is the spherical ellipse $\widehat{PA} + \widehat{PB} = \pi$, as stated above.
Thus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle $\Gamma$ the foci can be any two points $A$ and $B$ which lie on the same great circle perpendicular to $\Gamma$, on the same side of $\Gamma$ and at equal distances from $\Gamma$. The equation of any such spherical ellipse is $\widehat{PA} + \widehat{PB} = \boxed{\pi}$. | algebraic | putnam (modified boxing) | Geometry Algebra | A "spherical ellipse" with foci $A, B$ on a given sphere is defined as the set of all points $P$ on the sphere such that $\widehat{PA} + \widehat{PB} = \text{constant}$. Here $\widehat{PA}$ denotes the shortest distance on the sphere between $P$ and $A$. Determine the entire class of real spherical ellipses which are circles. | We take the radius of the sphere as unity and denote the constant sum $\widehat{PA} + \widehat{PB}$ by $2a$. To avoid trivial and degenerate cases we assume that $0 < \widehat{AB} < \pi$ and that $\widehat{AB} < 2a < 2\pi - \widehat{AB}$.
The case $2a > \pi$ can be reduced to the case $2a < \pi$. For, if $A'$ and $B'$ are the points diametrically opposite to $A$ and $B$ then $\widehat{PA} + \widehat{PB} = 2a$ if and only if $\widehat{PA}' + \widehat{PB}' = 2\pi - 2a$; that is, the spherical ellipses $\widehat{PA} + \widehat{PB} = 2a$ and $\widehat{PA}' + \widehat{PB}' = 2\pi - 2a$ are identical. Since $\min(2a, 2\pi - 2a) \leq \pi$, we may assume without loss of generality that $2a \leq \pi$.
Let $A$ and $B$ lie on the equator. There are two points $V_1$ and $V_2$ (the "vertices") on the equator which lie on the spherical ellipse. Obviously, $V_1V_2 = 2a$. The "center" of the spherical ellipse (common midpoint of the arcs $\widehat{AB}$ and $\widehat{V_1V_2}$) will be denoted by $C$.
We first treat the case $2a < \pi$ and show that in this case the spherical ellipse cannot be a circle. Assume it were a circle; call it $\Gamma$ (see Figure 1). $\Gamma$ would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. $\Gamma$ would also have to pass through the vertices. Therefore its spherical diameter would be $V_1V_2 = 2a$ and its spherical radius would be equal to $a$. The spherical center of $\Gamma$, the center of the ellipse, let $M$ be one of the two points on $\Gamma$ which lie half-way between the two vertices. Then, since $M$ is supposed to be a point on the spherical ellipse, $2a = \widehat{MA} + \widehat{MB} > \widehat{MC} = 2a$ (note that $\widehat{MC}$ is a right spherical triangle with the right angle at $C$ and with side $\widehat{MC} = a < 2a$). Contradiction shows that the only possible spherical ellipses which are circles must occur when $2a = \pi$.
In case $2a = \pi$, $V_1$ and $V_2$ are diametrically opposite points on the equator. We shall show that the great circle $\Gamma$ through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse $\widehat{PA} + \widehat{PB} = \pi$. To see this, let $B^*$ be the reflection of $B$ about the plane of $\Gamma$. $B^*$ is on the equator diametrically opposite to $A$ (see Fig. 2). Let $P$ be an arbitrary point on the sphere, and draw the great circle through $A$, $P$ and $B^*$. Then $\widehat{PA} + \widehat{PB} = \pi$ if and only if $\widehat{PB} = \widehat{PB^*}$; that is, if and only if $P$ is on $\Gamma$. This shows that $\Gamma$ is the spherical ellipse $\widehat{PA} + \widehat{PB} = \pi$, as stated above.
Thus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle $\Gamma$ the foci can be any two points $A$ and $B$ which lie on the same great circle perpendicular to $\Gamma$, on the same side of $\Gamma$ and at equal distances from $\Gamma$. The equation of any such spherical ellipse is $\boxed{\widehat{PA} + \widehat{PB} = \pi}$. | 0 |
1971 | 1971_B5 | Show that the graphs in the $x$-$y$ plane of all solutions of the system of differential equations \[x'' + y' + 6x = 0, \quad y'' - x' + 6y = 0 \quad (\,' = d/dt)\] which satisfy $x'(0) = y'(0) = 0$ are hypocycloids, and find the sum of the radius of the fixed circle and the two possible values of the radius of the rolling circle for each such solution. (A hypocycloid is the path described by a fixed point on the circumference of a circle which rolls on the inside of a given fixed circle.) | We put $z = x + iy$. Then both differential equations can be combined into one, namely\[z'' - iz' + 6z = 0.\]This is a standard linear equation of the second order with constant coefficients and has the general solution\[z(t) = c_1e^{3it} + c_2e^{-2it}.\]The initial conditions imply $z'(0) = 0$ or $3ic_1 - 2ic_2 = 0$. We may set $c_1 = 2A$ and $c_2 = 3A$, where $A$ is any complex number. The general solution of the given system is\[z(t) = 2Ae^{3it} + 3Ae^{-2it}.\]If $A = Re^{i\alpha}$, then a rotation of axes through the angle $\alpha$ produces\[Z(t) = 2Re^{3it} + 3Re^{-2it}\]or in rectangular form\[X(t) = 2R \cos (3t) + 3R \cos (2t)\\ Y(t) = 2R \sin (3t) - 3R \sin (2t).\]This is the standard form for a hypocycloid when the radius of the rolling circle is $3R$ and the fixed circle is of radius $5R$. On time reversal it becomes the standard equations of a hypocycloid with radius of the rolling circle of $2R$ and the radius of the fixed circle of $5R$.The final answer thus becomes \boxed{10R}. | algebraic | putnam (modified boxing) | Differential Equations Geometry | Show that the graphs in the $x$-$y$ plane of all solutions of the system of differential equations \[x'' + y' + 6x = 0, \quad y'' - x' + 6y = 0 \quad (\,' = d/dt)\] which satisfy $x'(0) = y'(0) = 0$ are hypocycloids, and find the radius of the fixed circle and the two possible values of the radius of the rolling circle for each such solution. (A hypocycloid is the path described by a fixed point on the circumference of a circle which rolls on the inside of a given fixed circle.) | We put $z = x + iy$. Then both differential equations can be combined into one, namely\[z'' - iz' + 6z = 0.\]This is a standard linear equation of the second order with constant coefficients and has the general solution\[z(t) = c_1e^{3it} + c_2e^{-2it}.\]The initial conditions imply $z'(0) = 0$ or $3ic_1 - 2ic_2 = 0$. We may set $c_1 = 2A$ and $c_2 = 3A$, where $A$ is any complex number. The general solution of the given system is\[z(t) = 2Ae^{3it} + 3Ae^{-2it}.\]If $A = Re^{i\alpha}$, then a rotation of axes through the angle $\alpha$ produces\[Z(t) = 2Re^{3it} + 3Re^{-2it}\]or in rectangular form\[X(t) = 2R \cos (3t) + 3R \cos (2t)\\ Y(t) = 2R \sin (3t) - 3R \sin (2t).\]This is the standard form for a hypocycloid when the \boxed{radius of the rolling circle is $3R$ and the fixed circle is of radius $5R$. On time reversal it becomes the standard equations of a hypocycloid with radius of the rolling circle of $2R$ and the radius of the fixed circle of $5R$}. | 0 |
1971 | 1971_B6 | Let $\delta(x)$ be the greatest odd divisor of the positive integer $x$. Find the upper bound of \left| \sum_{n=1}^x \frac{\delta(n)}{n} - \frac{2x}{3} \right| for all positive integers $x$. | Set \[ S(x) = \sum_{n=1}^x \frac{\delta(n)}{n}. \] Note that $\delta(2m+1) = 2m+1$, $\delta(2m) = \delta(m)$, and that $S(2x+1) = S(2x) + 1$. Dividing the summation for $S(2x)$ into even and odd values of the index produces the following relation: \[ S(2x) = \sum_{m=1}^x \frac{\delta(2m)}{2m} + \sum_{m=1}^x \frac{\delta(2m-1)}{2m-1} = \frac{1}{2}S(x) + x. \] If we denote $S(x) - \frac{2x}{3}$ by $F(x)$, the above relations translate into \[ F(2x) = \frac{1}{2}F(x), \quad \text{and} \quad F(2x+1) = F(2x) + \frac{1}{3}. \] Now induction can be used to show that $0 < F(x) < \boxed{\frac{2}{3}}$, for all positive integers $x$. This result is sharper than that requested. | numerical | putnam (modified boxing) | Number Theory Analysis | Let $\delta(x)$ be the greatest odd divisor of the positive integer $x$. Show that \[ \left| \sum_{n=1}^x \frac{\delta(n)}{n} - \frac{2x}{3} \right| < 1, \] for all positive integers $x$. | Set \[ S(x) = \sum_{n=1}^x \frac{\delta(n)}{n}. \] Note that $\delta(2m+1) = 2m+1$, $\delta(2m) = \delta(m)$, and that $S(2x+1) = S(2x) + 1$. Dividing the summation for $S(2x)$ into even and odd values of the index produces the following relation: \[ S(2x) = \sum_{m=1}^x \frac{\delta(2m)}{2m} + \sum_{m=1}^x \frac{\delta(2m-1)}{2m-1} = \frac{1}{2}S(x) + x. \] If we denote $S(x) - \frac{2x}{3}$ by $F(x)$, the above relations translate into \[ F(2x) = \frac{1}{2}F(x), \quad \text{and} \quad F(2x+1) = F(2x) + \frac{1}{3}. \] Now induction can be used to show that $0 < F(x) < \frac{2}{3}$, for all positive integers $x$. This result is sharper than that requested. | 0 |
1972 | 1972_B2 | A particle moving on a straight line starts from rest and attains a velocity $v_0$ after traversing a distance $s_0$. If the motion is such that the acceleration was never increasing, find the maximum time for the traverse. | We take $v_0$ as positive (see Comment) and consider the graph of $v$ as a function of $t$ (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration $a = dv/dt$ does not increase.
\[ \text{(See Figure for graphical representation)} \]
Let $t_0$ be the time of the traverse. Then $v(t_0) = v_0$. The distance $s_0$ is represented by the area bounded by the curve $v = v(t)$, the $t$-axis, and the line $t = t_0$. The area of the right triangle with vertices at $(0,0)$, $(t_0, 0)$ and $(t_0, v_0)$ has area less than or equal to $s$. Thus \[ \frac{1}{2} v_0 t_0 \leq s_0 \] or \[ t_0 \leq \boxed{\frac{2s_0}{v_0}}. \]
Equality is possible and gives the maximum value of $t_0$ (for given $s_0$ and $v_0$) when the graph of $v(t)$ is the straight line $v(t) = (v_0/t_0)t = (v_0^2 / 2s_0)t$. | algebraic | putnam | Differential Equations Calculus | A particle moving on a straight line starts from rest and attains a velocity $v_0$ after traversing a distance $s_0$. If the motion is such that the acceleration was never increasing, find the maximum time for the traverse. | We take $v_0$ as positive (see Comment) and consider the graph of $v$ as a function of $t$ (see Figure 1). From the given data we know that the curve starts at the origin and is concave downward since the acceleration $a = dv/dt$ does not increase.
\[ \text{(See Figure for graphical representation)} \]
Let $t_0$ be the time of the traverse. Then $v(t_0) = v_0$. The distance $s_0$ is represented by the area bounded by the curve $v = v(t)$, the $t$-axis, and the line $t = t_0$. The area of the right triangle with vertices at $(0,0)$, $(t_0, 0)$ and $(t_0, v_0)$ has area less than or equal to $s$. Thus \[ \frac{1}{2} v_0 t_0 \leq s_0 \] or \[ t_0 \leq \boxed{\frac{2s_0}{v_0}}. \]
Equality is possible and gives the maximum value of $t_0$ (for given $s_0$ and $v_0$) when the graph of $v(t)$ is the straight line $v(t) = (v_0/t_0)t = (v_0^2 / 2s_0)t$. | 0 |
1972 | 1972_B3 | Let $A$ and $B$ be two elements in a group such that $ABA = BA^2B$, $A^3 = 1$ and $B^{2n-1} = 1$ for some positive integer $n$. Find the value of $B$. | From $ABA = BA^2B = BA^{-1}B$, we have \[ AB^2 = ABA \cdot A^{-1}B = BA^{-1}BA^{-1}B = BA^{-1} \cdot ABA = B^2A. \] By induction, \[ AB^{2r} = B^{2r}A \] so that \[ AB = AB^{2n} = B^{2n}A = BA. \] Since $A$ and $B$ commute, $ABA = BA^2B$ implies \[ A^2B = A^2B^2, \] or \[ B = B^2, \] or \[ B = \boxed{1}. \] | numerical | putnam (modified boxing) | Algebra | Let $A$ and $B$ be two elements in a group such that $ABA = BA^2B$, $A^3 = 1$ and $B^{2n-1} = 1$ for some positive integer $n$. Prove $B = 1$. | From $ABA = BA^2B = BA^{-1}B$, we have \[ AB^2 = ABA \cdot A^{-1}B = BA^{-1}BA^{-1}B = BA^{-1} \cdot ABA = B^2A. \] By induction, \[ AB^{2r} = B^{2r}A \] so that \[ AB = AB^{2n} = B^{2n}A = BA. \] Since $A$ and $B$ commute, $ABA = BA^2B$ implies \[ A^2B = A^2B^2, \] or \[ B = B^2, \] or \[ B = 1. \] | 0 |
1973 | 1973_A1 | Let $ABC$ be any triangle. Let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively. Suppose the distances $BX \leq XC, CY \leq YA, AZ \leq ZB$. Given that the area of the triangle $XYZ$ is $\geq n$ (area of triangle $ABC$) find the lowest non-zero value of $n$. | (a) If $X, Y, Z$ are at the midpoints of the sides, the area of $\triangle XYZ$ is one fourth of the area of $\triangle ABC$. Also, as long as $BX \leq XC, CY \leq YA$ and $AZ \leq ZB$, moving one of $X, Y, Z$ to the midpoint of its side, while leaving the other two fixed, does not increase the area of $\triangle XYZ$ since the altitude to the fixed base of $\triangle XYZ$ decreases or remains constant. This gives us \boxed{\frac{1}{4}}. | numerical | putnam | Geometry | (a) Let $ABC$ be any triangle. Let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively. Suppose the distances $BX \leq XC, CY \leq YA, AZ \leq ZB$ (see Figure 1). Show that the area of the triangle $XYZ$ is $\geq (1/4)$ (area of triangle $ABC$).
(b) Let $ABC$ be any triangle, and let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively (but without any assumption about the ratios of the distances $BX/XC$, etc.; see Figures 1 and 2). Using (a) or by any other method, show: One of the three corner triangles $AZY, BXZ, CYX$ has an area $\leq$ area of triangle $XYZ$. | (a) If $X, Y, Z$ are at the midpoints of the sides, the area of $\triangle XYZ$ is one fourth of the area of $\triangle ABC$. Also, as long as $BX \leq XC, CY \leq YA$ and $AZ \leq ZB$, moving one of $X, Y, Z$ to the midpoint of its side, while leaving the other two fixed, does not increase the area of $\triangle XYZ$ since the altitude to the fixed base of $\triangle XYZ$ decreases or remains constant.
(b) Under the hypothesis of (a) the three corner triangles have no more than three fourths of the total area and so one of them must have smaller area than $\triangle XYZ$. All other cases are similar to the one in which $XC < BX$ and $CY < YA$. Then consideration of the altitudes to base $XY$ shows that $\triangle CYX$ has smaller area than $\triangle XYZ$. | 0 |
1973 | 1973_A4 | How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line? [By a "zero" of a function $f$, we mean a value $x_0$ in the domain of $f$ (here the set of all real numbers) such that $f(x_0) = 0.] | Three; at $0$, $1$, and some $x > 1$. The first two are clear and the other follows from $f(4) < 0$ and $f(5) > 0$ or from $f'(1) < 0$ while $f(x) \to +\infty$ as $x \to +\infty$. There are no more zeros since four zeros of $f$ would imply a zero of $f''$ using an extension of Rolle's Theorem; but $f''(x) = (\log 2)^3 2^x \neq 0$ for all $x$. Thus making the number \boxed{3}. | numerical | putnam | Analysis | How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line? [By a "zero" of a function $f$, we mean a value $x_0$ in the domain of $f$ (here the set of all real numbers) such that $f(x_0) = 0.] | \boxed{Three}; at $0$, $1$, and some $x > 1$. The first two are clear and the other follows from $f(4) < 0$ and $f(5) > 0$ or from $f'(1) < 0$ while $f(x) \to +\infty$ as $x \to +\infty$. There are no more zeros since four zeros of $f$ would imply a zero of $f''$ using an extension of Rolle's Theorem; but $f''(x) = (\log 2)^3 2^x \neq 0$ for all $x$. | 0 |
1973 | 1973_B3 | Consider an integer $p > 1$ with the property that the polynomial $x^2 - x + p$ takes prime values for all integers $x$ in the range $0 \leq x < p$. (Examples: $p = 5$ and $p = 41$ have this property.) Find triples of integers $a, b, c$ satisfying the conditions:
\[
b^2 - 4ac = 1 - 4p,\]
\[
0 < a \leq c,\]
\[
-a \leq b < a.
\] | One triple $(a, b, c)$ satisfying the conditions is $(1, -1, p)$; it remains to show that this is the only solution. Clearly $b$ must be odd since $b^2 \equiv 1 \pmod{4}$. Also $b^2 = (-b)^2$, so write $|b| = 2x - 1$. Then $b^2 - 4ac = 1 - 4p$ gives
\[x^2 - x + p = ac.\]
If $0 \leq x < p$, the hypothesis tells us that $ac$ is prime; then $0 < a \leq c$ implies that $a = 1$, it follows from $-a \leq b < a$ and the oddness of $b$ that $b = -1$, and $1 - 4p = b^2 - 4ac = 1 - 4p$ gives us $c = p$. Since $x = (|b| + 1)/2 \geq 0$, it suffices to show that $x < p$. Since
\[3a^2 = 4a^2 - a^2 \leq 4ac - b^2 = 4p - 1,\]
\[|b| \leq a \leq \sqrt{(4p - 1)/3},\]
\[x = (|b| + 1)/2 < \sqrt{p/3} + (1/2) < p.\] Thus the only solution is \boxed{(1, -1, p)}. | algebraic | putnam (modified boxing) | Number Theory Algebra | Consider an integer $p > 1$ with the property that the polynomial $x^2 - x + p$ takes prime values for all integers $x$ in the range $0 \leq x < p$. (Examples: $p = 5$ and $p = 41$ have this property.) Show that there is exactly one triple of integers $a, b, c$ satisfying the conditions:
\[
b^2 - 4ac = 1 - 4p,\]
\[
0 < a \leq c,\]
\[
-a \leq b < a.
\] | One triple $(a, b, c)$ satisfying the conditions is $(1, -1, p)$; it remains to show that this is the only solution. Clearly $b$ must be odd since $b^2 \equiv 1 \pmod{4}$. Also $b^2 = (-b)^2$, so write $|b| = 2x - 1$. Then $b^2 - 4ac = 1 - 4p$ gives
\[x^2 - x + p = ac.\]
If $0 \leq x < p$, the hypothesis tells us that $ac$ is prime; then $0 < a \leq c$ implies that $a = 1$, it follows from $-a \leq b < a$ and the oddness of $b$ that $b = -1$, and $1 - 4p = b^2 - 4ac = 1 - 4p$ gives us $c = p$. Since $x = (|b| + 1)/2 \geq 0$, it suffices to show that $x < p$. Since
\[3a^2 = 4a^2 - a^2 \leq 4ac - b^2 = 4p - 1,\]
\[|b| \leq a \leq \sqrt{(4p - 1)/3},\]
\[x = (|b| + 1)/2 < \sqrt{p/3} + (1/2) < p.\] | 0 |
1973 | 1973_B6 | On the domain $0 \leq \theta \leq 2\pi$ given that \[ \left| \sin^2\theta \{ \sin^3(2\theta) \cdot \sin^3(4\theta) \cdots \sin^3(2^{n-1}\theta) \} \sin(2^n\theta) \right| \] find the minimum positive $\theta$ value at which the value is maximized. (The maximum may also be attained at other points.) | By induction: The case $n = 1$ is just simple calculus. Now the ratio of the expression for $n+1$ to the expression for $n$ is equal to: \[ \left| \sin^2(2^n\theta) \cdot \sin(2^{n+1}\theta) \right|. \] Since $\theta = \pi/3$ gives $2^n\theta \equiv 2\pi/3$ or $4\pi/3 \pmod{2\pi}$, this ratio is maximized at $\theta = \boxed{\pi/3}$, and by induction, then, the whole expression is maximized. | numerical | putnam (modified boxing) | Trigonometry Analysis | On the domain $0 \leq \theta \leq 2\pi$: \begin{enumerate} \item[(a)] Prove that $\sin^2\theta \cdot \sin(2\theta)$ takes its maximum at $\pi/3$ and $4\pi/3$ (and hence its minimum at $2\pi/3$ and $5\pi/3$). \item[(b)] Show that \[ \left| \sin^2\theta \{ \sin^3(2\theta) \cdot \sin^3(4\theta) \cdots \sin^3(2^{n-1}\theta) \} \sin(2^n\theta) \right| \] takes its maximum at $\theta = \pi/3$. (The maximum may also be attained at other points.) \item[(c)] Derive the inequality: \[ \sin^2\theta \cdot \sin^2(2\theta) \cdot \sin^2(4\theta) \cdots \sin^2(2^n\theta) \leq (3/4)^n. \] \end{enumerate} | \begin{enumerate} \item[(a)] Simple calculus. \item[(b)] By induction: The case $n = 1$ is just (a). \newline Now the ratio of the expression for $n+1$ to the expression for $n$ is equal to: \[ \left| \sin^2(2^n\theta) \cdot \sin(2^{n+1}\theta) \right|. \] Since $\theta = \pi/3$ gives $2^n\theta \equiv 2\pi/3$ or $4\pi/3 \pmod{2\pi}$, this ratio is maximized at $\theta = \pi/3$, and by induction, then, the whole expression is maximized. \item[(c)] Set $\theta = \pi/3$, and observe that the expression in part (b) is then exactly equal to $(3/4)^{n/2}$; its $2/3$ power is thus equal to $(3/4)^n$. That is the maximum; in general the $2/3$ power of the expression in (b) is $\leq (3/4)^n$. To get from that to the expression in (c), we would increase the powers of the end factors $\sin\theta$ and $\sin(2^n\theta)$; this can only decrease the product, since $|\sin\theta| \leq 1$. \end{enumerate} | 0 |
1974 | 1974_A1 | Call a set of positive integers "conspiratorial" if no three of them are pairwise relatively prime. (A set of integers is "pairwise relatively prime" if no pair of them has a common divisor greater than 1.) What is the largest number of elements in any "conspiratorial" subset of the integers 1 through 16? | A conspiratorial subset (CS) of \{1, 2, \dots, 16\} has at most two numbers from the pairwise relatively prime set \{1, 2, 3, 5, 7, 11, 13\} and so has at most $16 - (7 - 2) = 11$ numbers. But \{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16\} is a CS with 11 elements; hence the answer is \boxed{11}. | numerical | putnam | Number Theory Combinatorics | Call a set of positive integers "conspiratorial" if no three of them are pairwise relatively prime. (A set of integers is "pairwise relatively prime" if no pair of them has a common divisor greater than 1.) What is the largest number of elements in any "conspiratorial" subset of the integers 1 through 16? | A conspiratorial subset (CS) of \{1, 2, \dots, 16\} has at most two numbers from the pairwise relatively prime set \{1, 2, 3, 5, 7, 11, 13\} and so has at most $16 - (7 - 2) = 11$ numbers. But \{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16\} is a CS with 11 elements; hence the answer is \boxed{11}. | 0 |
1974 | 1974_A3 | A well-known theorem asserts that a prime $p > 2$ can be written as the sum of two perfect squares ($p = m^2 + n^2$, with $m$ and $n$ integers) if and only if $p \equiv 1 \pmod{4}$. Assuming this result, find which primes $p > 2$ can be written in each of the following forms, using (not necessarily positive) integers $x$ and $y$: \begin{enumerate} \item[(a)] $x^2 + 16y^2;$ \item[(b)] $4x^2 + 4xy + 5y^2$. \end{enumerate} Give the answer in the form of an an equivalance expression. | If $p \equiv 1 \pmod{4}$, either (A): $p \equiv 1 \pmod{8}$ or (B): $p \equiv 5 \pmod{8}$. We show that (A) and (B) are necessary and sufficient for (a) and (b), respectively. If $p = m^2 + n^2$ and $p$ is odd, one can let $m$ be odd and $n$ be even. Then $p = m^2 + 4v^2$ with $m^2 \equiv 1 \pmod{8}$. With (A), $v$ is even and $p = m^2 + 16w^2$ implies $p \equiv m^2 \equiv 1 \pmod{8}$. With (B), $v$ is odd, $m = 2u + v$ for some integer $u$, and $p = (2u + v)^2 + 4v^2 = 4u^2 + 4uv + 5v^2$. Conversely, $p = 4u^2 + 4uv + 5v^2$ with $p$ odd implies $p = (2u + v)^2 + 4v^2$ with $v$ odd and hence $\boxed{p \equiv 5 \pmod{8}}$. | algebraic | putnam | Number Theory | A well-known theorem asserts that a prime $p > 2$ can be written as the sum of two perfect squares ($p = m^2 + n^2$, with $m$ and $n$ integers) if and only if $p \equiv 1 \pmod{4}$. Assuming this result, find which primes $p > 2$ can be written in each of the following forms, using (not necessarily positive) integers $x$ and $y$: \begin{enumerate} \item[(a)] $x^2 + 16y^2;$ \item[(b)] $4x^2 + 4xy + 5y^2$. \end{enumerate} | If $p \equiv 1 \pmod{4}$, either (A): $p \equiv 1 \pmod{8}$ or (B): $p \equiv 5 \pmod{8}$. We show that (A) and (B) are necessary and sufficient for (a) and (b), respectively. If $p = m^2 + n^2$ and $p$ is odd, one can let $m$ be odd and $n$ be even. Then $p = m^2 + 4v^2$ with $m^2 \equiv 1 \pmod{8}$. With (A), $v$ is even and $p = m^2 + 16w^2$ implies $p \equiv m^2 \equiv 1 \pmod{8}$. With (B), $v$ is odd, $m = 2u + v$ for some integer $u$, and $p = (2u + v)^2 + 4v^2 = 4u^2 + 4uv + 5v^2$. Conversely, $p = 4u^2 + 4uv + 5v^2$ with $p$ odd implies $p = (2u + v)^2 + 4v^2$ with $v$ odd and hence $\boxed{p \equiv 5 \pmod{8}}$. | 0 |
1974 | 1974_A4 | An unbiased coin is tossed $n$ times. What is the expected value of $|H - T|$, where $H$ is the number of heads and $T$ is the number of tails? In other words, evaluate in \textit{closed form}:
\[\frac{1}{2^{n-1}} \sum_{k < n/2} (n-2k)\binom{n}{k}.\] | Since
\[ \sum_{k < n/2} (n-2k) \binom{n}{k} = \sum_{k < n/2} \{(n-k)\binom{n}{k} - k\binom{n}{k}\}
= \sum_{k < n/2} \{n\binom{n-1}{k} - n\binom{n-1}{k-1}\} = n\sum_{k < n/2} \binom{n-1}{k} - \binom{n-1}{k-1} \}
= n\binom{n-1}{\lfloor (n-1)/2 \rfloor}.\]The answer is
\[\boxed{\frac{n}{2^{n-1}}\binom{n-1}{\lfloor (n-1)/2 \rfloor}}\]. | algebraic | putnam | Probability Combinatorics | An unbiased coin is tossed $n$ times. What is the expected value of $|H - T|$, where $H$ is the number of heads and $T$ is the number of tails? In other words, evaluate in \textit{closed form}:
\[\frac{1}{2^{n-1}} \sum_{k < n/2} (n-2k)\binom{n}{k}.\]
(In this problem, "closed form" means a form not involving a series. The given series can be reduced to a single term involving only binomial coefficients, rational functions of $n$ and $2^n$, and the greatest integer function $\lfloor x \rfloor$.) | The answer is
\[\boxed{\frac{n}{2^{n-1}}\binom{n-1}{\lfloor (n-1)/2 \rfloor}}\]
since
\[ \sum_{k < n/2} (n-2k) \binom{n}{k} = \sum_{k < n/2} \{(n-k)\binom{n}{k} - k\binom{n}{k}\}
= \sum_{k < n/2} \{n\binom{n-1}{k} - n\binom{n-1}{k-1}\} = n\sum_{k < n/2} \binom{n-1}{k} - \binom{n-1}{k-1} \}
= n\binom{n-1}{\lfloor (n-1)/2 \rfloor}.\] | 0 |
1974 | 1974_A5 | Consider the two mutually tangent parabolas $y = x^2$ and $y = -x^2$. [These have foci at $(0, 1/4)$ and $(0, -1/4)$, and directrices $y = -1/4$ and $y = 1/4$, respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola. | Let $F$ be the fixed focus, $M$ be the moving focus, and $T$ be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at $T$ makes equal angles with $FT$ and with a vertical line. This and congruence of the two parabolas imply that $MT$ is vertical and that the segments $FT$ and $MT$ are equal. Now $M$ must be on the horizontal fixed directrix $\boxed{y = 1/4}$ by the focus-directrix definition of a parabola. | algebraic | putnam | Geometry | Consider the two mutually tangent parabolas $y = x^2$ and $y = -x^2$. [These have foci at $(0, 1/4)$ and $(0, -1/4)$, and directrices $y = -1/4$ and $y = 1/4$, respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola. | Let $F$ be the fixed focus, $M$ be the moving focus, and $T$ be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at $T$ makes equal angles with $FT$ and with a vertical line. This and congruence of the two parabolas imply that $MT$ is vertical and that the segments $FT$ and $MT$ are equal. Now $M$ must be on the horizontal fixed directrix $\boxed{y = 1/4}$ by the focus-directrix definition of a parabola. | 0 |
1974 | 1974_A6 | It is well known that the value of the polynomial $(x+1)(x+2)\cdots(x+n)$ is exactly divisible by $n$ for every integer $x$. Given $n$, let $k=k(n)$ be the minimal degree of any monic integral polynomial
\[ f(x) = x^k + a_{k-1}x^{k-1} + \cdots + a_k \]
(with integer coefficients and leading coefficient $1$) such that the value of $f(x)$ is exactly divisible by $n$ for every integer $x$.
Find the relationship between $n$ and $k=k(n)$. In particular, find the value of $k$ corresponding to $n=1,000,000.$ | Let $p(k, x)$ be the monic polynomial $(x+1)(x+2)\cdots(x+k)$ and let $m$ be an integer. Then $p(k, m)$ is exactly divisible by $k!$ since the absolute value of the quotient is a binomial coefficient (even when $m$ is negative). Hence, if $n\mid k!$ there is a monic integral polynomial $f(x)$ of degree $k$ with $n\mid f(m)$ for all integers $m$. Conversely, the condition $n\mid k!$ is necessary since the $k$-th difference $k!$ of a monic integral polynomial of degree $k$ is divisible by any common divisor of all the values $f(m)$.
In particular, $k(10^6) = k(5^6 2^6) = 25$ since the smallest $s$ with $5^6\mid s!$ is $s=\boxed{25}.$ | numerical | putnam | Algebra Number Theory | It is well known that the value of the polynomial $(x+1)(x+2)\cdots(x+n)$ is exactly divisible by $n$ for every integer $x$. Given $n$, let $k=k(n)$ be the minimal degree of any monic integral polynomial
\[ f(x) = x^k + a_{k-1}x^{k-1} + \cdots + a_k \]
(with integer coefficients and leading coefficient $1$) such that the value of $f(x)$ is exactly divisible by $n$ for every integer $x$.
Find the relationship between $n$ and $k=k(n)$. In particular, find the value of $k$ corresponding to $n=1,000,000.$ | Let $p(k, x)$ be the monic polynomial $(x+1)(x+2)\cdots(x+k)$ and let $m$ be an integer. Then $p(k, m)$ is exactly divisible by $k!$ since the absolute value of the quotient is a binomial coefficient (even when $m$ is negative). Hence, if $n\mid k!$ there is a monic integral polynomial $f(x)$ of degree $k$ with $n\mid f(m)$ for all integers $m$. Conversely, the condition $n\mid k!$ is necessary since the $k$-th difference $k!$ of a monic integral polynomial of degree $k$ is divisible by any common divisor of all the values $f(m)$.
In particular, $k(10^6) = k(5^6 2^6) = 25$ since the smallest $s$ with $5^6\mid s!$ is $s=\boxed{25}.$ | 0 |
1974 | 1974_B6 | For a set with $n$ elements, how many subsets are there whose cardinality (the number of elements in the subset) is respectively $\equiv 0 \pmod{3}, \equiv 1 \pmod{3}, \equiv 2 \pmod{3}$? In other words, calculate \[ s_{i,n} = \sum_{k \equiv i \pmod{3}} \binom{n}{k} \quad \text{for } i = 0, 1, 2 and any given $n$. \] | Let $n \equiv r \pmod{6}$ with $r \in \{0,1,2,3,4,5\}$. Then the pattern is \[ \begin{aligned} r & \quad 0 \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \\ s_{0,n} & \quad a+1 \quad b \quad c \quad d-1 \quad e \quad f \\ s_{1,n} & \quad a \quad b \quad c+1 \quad d \quad e \quad f-1 \\ s_{2,n} & \quad a \quad b-1 \quad c \quad d \quad e+1 \quad f \end{aligned} \] This is easily proved by mathematical induction using the formulas \[ s_{i,n} = s_{i-1,n-1} + s_{i,n-1}. \quad \text{[Here } 0-1 = 2 \pmod{3}\text{.]} \] These formulas follow immediately from the rule \[ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \] The sums may be computed readily using the above patterns and \[ s_{0,n} + s_{1,n} + s_{2,n} = 2^n. \] For $n = 1000$, $r = 4$ and \[ s_{0,1000} = s_{1,1000} = s_{2,1000} - 1 = 2^{1000} - 1}{3}. Thus making the sum for any $n$, \boxed{2^n}. \] | numerical | putnam (modified boxing) | Combinatorics | For a set with $n$ elements, how many subsets are there whose cardinality (the number of elements in the subset) is respectively $\equiv 0 \pmod{3}, \equiv 1 \pmod{3}, \equiv 2 \pmod{3}$? In other words, calculate \[ s_{i,n} = \sum_{k \equiv i \pmod{3}} \binom{n}{k} \quad \text{for } i = 0, 1, 2. \] | Let $n \equiv r \pmod{6}$ with $r \in \{0,1,2,3,4,5\}$. Then the pattern is \[ \begin{aligned} r & \quad 0 \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \\ s_{0,n} & \quad a+1 \quad b \quad c \quad d-1 \quad e \quad f \\ s_{1,n} & \quad a \quad b \quad c+1 \quad d \quad e \quad f-1 \\ s_{2,n} & \quad a \quad b-1 \quad c \quad d \quad e+1 \quad f \end{aligned} \] This is easily proved by mathematical induction using the formulas \[ s_{i,n} = s_{i-1,n-1} + s_{i,n-1}. \quad \text{[Here } 0-1 = 2 \pmod{3}\text{.]} \] These formulas follow immediately from the rule \[ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \] The sums may be computed readily using the above patterns and \[ s_{0,n} + s_{1,n} + s_{2,n} = 2^n. \] For $n = 1000$, $r = 4$ and \[ s_{0,1000} = s_{1,1000} = s_{2,1000} - 1 = \boxed{\frac{2^{1000} - 1}{3}}. \] | 0 |
1975 | 1975_A2 | For which ordered pairs of real numbers $b, c$ do both roots of the quadratic equation \[ z^2 + bz + c = 0 \] lie inside the unit disk $\{ |z| < 1 \}$ in the complex plane? Find the sum of corresponding values in the ordered pairs and return the final summed ordered pair. | The desired region is the inside of the triangle with vertices $(0, -1)$, $(2, 1)$, $(-2, 1)$. The boundary segments lie on the lines \[ L_1: c = 1, \quad L_2: c - b + 1 = 0, \quad L_3: c + b + 1 = 0. \] To see this, we let $f(z) = z^2 + bz + c$ and denote its zeros by $r$ and $s$. Then $-b = r + s$ and $c = rs$. Also \[ (r+1)(s+1) = rs + r + s + 1 = c - b + 1 = f(-1), \] \[ (r-1)(s-1) = rs - r - s + 1 = c + b + 1 = f(1). \] On or below $L_2$, at least one zero is real and not greater than $-1$; this follows either from $(r+1)(s+1) \leq 0$ or from $f(-1) \leq 0$ and the fact that the graph of $y = f(x)$, for $x$ real, is an upward-opening parabola. Similarly, on or below $L_3$, one zero is real and at least 1. On or above $L_1$, at least one zero has absolute value greater than or equal to 1. Hence the desired points $(b, c)$ must be inside the described triangle. Conversely, if $(b, c)$ is inside the triangle, $|c| < 1$ and so $|r| < 1$ or $|s| < 1$ or both. If the zeros are complex, they are conjugates and $|r| = |s|$; then $|r| = |s| < 1$ follows from $|c| < 1$. If the zeros are real, $|c| < 1$ implies that at least one zero is in $(-1, 1)$. Then $(r+1)(s+1) = f(-1) > 0$ and $(r-1)(s-1) = f(1) > 0$ imply that the other zero is also in $(-1, 1)$. The final sum is \boxed{(0, 1)}. | numerical | putnam (modified boxing) | Algebra Complex Numbers | For which ordered pairs of real numbers $b, c$ do both roots of the quadratic equation \[ z^2 + bz + c = 0 \] lie inside the unit disk $\{ |z| < 1 \}$ in the complex plane? Draw a reasonably accurate picture (i.e., 'graph') of the region in the real $bc$-plane for which the above condition holds. Identify precisely the boundary curves of this region. | The desired region is the inside of the triangle with vertices $\boxed{(0, -1)$, $(2, 1)$, $(-2, 1)}$. The boundary segments lie on the lines \[ L_1: c = 1, \quad L_2: c - b + 1 = 0, \quad L_3: c + b + 1 = 0. \] To see this, we let $f(z) = z^2 + bz + c$ and denote its zeros by $r$ and $s$. Then $-b = r + s$ and $c = rs$. Also \[ (r+1)(s+1) = rs + r + s + 1 = c - b + 1 = f(-1), \] \[ (r-1)(s-1) = rs - r - s + 1 = c + b + 1 = f(1). \] On or below $L_2$, at least one zero is real and not greater than $-1$; this follows either from $(r+1)(s+1) \leq 0$ or from $f(-1) \leq 0$ and the fact that the graph of $y = f(x)$, for $x$ real, is an upward-opening parabola. Similarly, on or below $L_3$, one zero is real and at least 1. On or above $L_1$, at least one zero has absolute value greater than or equal to 1. Hence the desired points $(b, c)$ must be inside the described triangle. Conversely, if $(b, c)$ is inside the triangle, $|c| < 1$ and so $|r| < 1$ or $|s| < 1$ or both. If the zeros are complex, they are conjugates and $|r| = |s|$; then $|r| = |s| < 1$ follows from $|c| < 1$. If the zeros are real, $|c| < 1$ implies that at least one zero is in $(-1, 1)$. Then $(r+1)(s+1) = f(-1) > 0$ and $(r-1)(s-1) = f(1) > 0$ imply that the other zero is also in $(-1, 1)$. For full credit, the region had to be depicted. | 0 |
1975 | 1975_A4 | Let $n = 2m$, where $m$ is an odd integer greater than 1. Let $\theta = e^{2\pi i / n}$. Express $(1-\theta)^{-1}$ explicitly as a polynomial in $\theta$,
$a_k \theta^k + a_{k-1} \theta^{k-1} + \cdots + a_1 \theta + a_0$,
with integer coefficients $a_i$.
[Note that $\theta$ is a primitive $n$-th root of unity, and thus it satisfies all of the identities which hold for such roots.] | Let $n = 4k + 2$ with $k > 0$. Then
$$0 = \theta^n - 1 = \theta^{4k+2} - 1 = (\theta^{2k+1} - 1)(\theta + 1)(\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots - \theta + 1).$$
Since $\theta$ is a primitive $n$-th root of unity with $n > 2k + 1$ and $n > 2$,
$$(\theta^{2k+1} - 1)(\theta + 1) \neq 0.$$
Hence
(A)
$$\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots + \theta^2 - \theta + 1 = 0,$$
$$1 = \theta - \theta^2 + \theta^3 - \cdots - \theta^{2k} = (1 - \theta)(\theta + \theta^2 + \theta^3 + \cdots + \theta^{2k-1}),$$
$$(1 - \theta)^{-1} = \theta + \theta^3 + \cdots + \theta^{2k-1} \quad \text{[where $2k - 1 = (n - 4)/2$]}.$$
Another solution is $(1 - \theta)^{-1} = \boxed{1 + \theta^2 + \theta^4 + \cdots + \theta^{2k}}$ as one sees from (A). | algebraic | putnam | Algebra Complex Numbers | Let $n = 2m$, where $m$ is an odd integer greater than 1. Let $\theta = e^{2\pi i / n}$. Express $(1-\theta)^{-1}$ explicitly as a polynomial in $\theta$,
$a_k \theta^k + a_{k-1} \theta^{k-1} + \cdots + a_1 \theta + a_0$,
with integer coefficients $a_i$.
[Note that $\theta$ is a primitive $n$-th root of unity, and thus it satisfies all of the identities which hold for such roots.] | Let $n = 4k + 2$ with $k > 0$. Then
$$0 = \theta^n - 1 = \theta^{4k+2} - 1 = (\theta^{2k+1} - 1)(\theta + 1)(\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots - \theta + 1).$$
Since $\theta$ is a primitive $n$-th root of unity with $n > 2k + 1$ and $n > 2$,
$$(\theta^{2k+1} - 1)(\theta + 1) \neq 0.$$
Hence
(A)
$$\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots + \theta^2 - \theta + 1 = 0,$$
$$1 = \theta - \theta^2 + \theta^3 - \cdots - \theta^{2k} = (1 - \theta)(\theta + \theta^2 + \theta^3 + \cdots + \theta^{2k-1}),$$
$$(1 - \theta)^{-1} = \theta + \theta^3 + \cdots + \theta^{2k-1} \quad \text{[where $2k - 1 = (n - 4)/2$]}.$$
Another solution is $(1 - \theta)^{-1} = \boxed{1 + \theta^2 + \theta^4 + \cdots + \theta^{2k}}$ as one sees from (A). | 0 |
1975 | 1975_B1 | In the additive group of ordered pairs of integers $(m,n)$ [with addition defined componentwise: $(m,n)+(m',n')=(m+m',n+n')$] consider the subgroup $H$ generated by the three elements \[(3,8), \quad (4,-1), \quad (5,4).\] Then $H$ has another set of generators of the form \[(1,b), \quad (0,a),\] for some integers $a,b$ with $a>0$. Find $a$. | Also one must have $b\equiv 5 \pmod{7}$. \textbf{Proof:} The subgroup $H$ must contain $4(3,8)-3(4,-1)=(0,35), \ 4(5,4)-5(4,-1)=(0,21)$, and then $2(0,21)-(0,35)=(0,7)$. Now $(0,7)$ and $(1,b)$ will generate $H$ iff $(1,b)$ is in $H$ and there exist integers $u,v,$ and $w$ such that \[(3,8)=3(1,b)+u(0,7), \quad (4,-1)=4(1,b)+v(0,7), \quad (5,4)=5(1,b)+w(0,7).\] These hold iff $8=3b+7u, \ -1=4b+7v, \ \text{and } 4=5b+7w.$ With $b=5+7k, \ k \text{ any integer, the desired coefficients } u,v, \text{ and } w \text{ exist in the form } u=-1-3k, \ v=-3-4k, \ w=-3-5k.$ It now suffices to let $k=0$ and to note that $(1,5)=(4,-1)-(3,8)+2(0,7) \text{ is in } H.$ The answer is $a=\boxed{7}$. | numerical | putnam | Algebra Number Theory | In the additive group of ordered pairs of integers $(m,n)$ [with addition defined componentwise: $(m,n)+(m',n')=(m+m',n+n')$] consider the subgroup $H$ generated by the three elements \[(3,8), \quad (4,-1), \quad (5,4).\] Then $H$ has another set of generators of the form \[(1,b), \quad (0,a),\] for some integers $a,b$ with $a>0$. Find $a$. \[\text{[Elements } g_1,\ldots,g_k \text{ are said to generate a subgroup } H \text{ if (i) each } g_i \in H, \text{ and (ii) every } h \in H \text{ can be written as a sum } h=n_1g_1+\cdots+n_kg_k \text{ where the } n_i \text{ are integers (and where, for example, } 3g_1-2g_2 \text{ means } g_1+g_1+g_1-g_2-g_2).\]} | The answer is $a=\boxed{7}$. Also one must have $b\equiv 5 \pmod{7}$. \textbf{Proof:} The subgroup $H$ must contain $4(3,8)-3(4,-1)=(0,35), \ 4(5,4)-5(4,-1)=(0,21)$, and then $2(0,21)-(0,35)=(0,7)$. Now $(0,7)$ and $(1,b)$ will generate $H$ iff $(1,b)$ is in $H$ and there exist integers $u,v,$ and $w$ such that \[(3,8)=3(1,b)+u(0,7), \quad (4,-1)=4(1,b)+v(0,7), \quad (5,4)=5(1,b)+w(0,7).\] These hold iff $8=3b+7u, \ -1=4b+7v, \ \text{and } 4=5b+7w.$ With $b=5+7k, \ k \text{ any integer, the desired coefficients } u,v, \text{ and } w \text{ exist in the form } u=-1-3k, \ v=-3-4k, \ w=-3-5k.$ It now suffices to let $k=0$ and to note that $(1,5)=(4,-1)-(3,8)+2(0,7) \text{ is in } H.$ | 0 |
1975 | 1975_B3 | Let $s_k(a_1, \ldots, a_n)$ denote the $k$-th elementary symmetric function of $a_1, \ldots, a_n$. With $k$ held fixed, find the supremum (or least upper bound) $M_k$ of \[ \frac{s_k(a_1, \ldots, a_n)}{[s_1(a_1, \ldots, a_n)]^k} \] for arbitrary $n \geq k$ and arbitrary $n$-tuples $a_1, \ldots, a_n$ of positive real numbers. \[ \text{[The symmetric function $s_k(a_1, \ldots, a_n)$ is the sum of all $k$-fold products of the variables $a_1, \ldots, a_n$. Thus, for example:]} \] \[ s_1(a_1, \ldots, a_n) = a_1 + a_2 + \cdots + a_n; \] \[ s_3(a_1, a_2, a_3, a_4) = a_1a_2a_3 + a_1a_3a_4 + a_2a_3a_4. \] \[ \text{It should be remarked that the supremum $M_k$ is never attained; it is approached arbitrarily closely when, for fixed $k$, the number $n$ of variables increases without bound, and the values $a_i > 0$ are suitably chosen.} \] | In the expansion of $s_k^1 = (a_1 + a_2 + \cdots + a_n)^k$, every term of $s_k$ appears with $k!$ as coefficient and the other coefficients are nonnegative. Hence \[ \frac{s_k}{s_1^k} \leq \frac{1}{k!} \] If we let each $a_i = 1$, \[ \frac{s_k}{s_1^k} = \frac{\binom{n}{k}}{n^k} = \frac{n(n-1)\cdots(n-k+1)}{k! \cdot n^k} = \frac{1}{k!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k-1}{n}\right), \] which approaches $\frac{1}{k!}$ as $k$ is held fixed and $n$ goes to infinity. These facts show that the supremum $M_k$ is $\boxed{\frac{1}{k!}}$. | algebraic | putnam | Algebra | Let $s_k(a_1, \ldots, a_n)$ denote the $k$-th elementary symmetric function of $a_1, \ldots, a_n$. With $k$ held fixed, find the supremum (or least upper bound) $M_k$ of \[ \frac{s_k(a_1, \ldots, a_n)}{[s_1(a_1, \ldots, a_n)]^k} \] for arbitrary $n \geq k$ and arbitrary $n$-tuples $a_1, \ldots, a_n$ of positive real numbers. \[ \text{[The symmetric function $s_k(a_1, \ldots, a_n)$ is the sum of all $k$-fold products of the variables $a_1, \ldots, a_n$. Thus, for example:]} \] \[ s_1(a_1, \ldots, a_n) = a_1 + a_2 + \cdots + a_n; \] \[ s_3(a_1, a_2, a_3, a_4) = a_1a_2a_3 + a_1a_3a_4 + a_2a_3a_4. \] \[ \text{It should be remarked that the supremum $M_k$ is never attained; it is approached arbitrarily closely when, for fixed $k$, the number $n$ of variables increases without bound, and the values $a_i > 0$ are suitably chosen.} \] | In the expansion of $s_k^1 = (a_1 + a_2 + \cdots + a_n)^k$, every term of $s_k$ appears with $k!$ as coefficient and the other coefficients are nonnegative. Hence \[ \frac{s_k}{s_1^k} \leq \frac{1}{k!} \] If we let each $a_i = 1$, \[ \frac{s_k}{s_1^k} = \frac{\binom{n}{k}}{n^k} = \frac{n(n-1)\cdots(n-k+1)}{k! \cdot n^k} = \frac{1}{k!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k-1}{n}\right), \] which approaches $\frac{1}{k!}$ as $k$ is held fixed and $n$ goes to infinity. These facts show that the supremum $M_k$ is $\boxed{\frac{1}{k!}}$. | 0 |
1975 | 1975_B5 | Let $f_0(x) = e^x$ and $f_n(x) = xf'_n(x)$ for $n = 0, 1, 2, \dots$ Find \sum_{n=0}^\infty \frac{f_n(1)}{n!} in terms of e. | Since $f_0(x) = \sum_{k=0}^\infty \frac{x^k}{k!}$, one easily shows by mathematical induction that \[ f_n(x) = \sum_{k=0}^\infty \frac{k^n x^k}{k!}. \] Then, since all terms are positive, one has \[ \sum_{n=0}^\infty \frac{f_n(1)}{n!} = \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{k^n}{k! n!} = \sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^\infty \frac{k^n}{n!} = \sum_{k=0}^\infty \frac{e^k}{k!} = \boxed{e^e}. \] | algebraic | putnam (modified boxing) | Calculus Analysis | Let $f_0(x) = e^x$ and $f_n(x) = xf'_n(x)$ for $n = 0, 1, 2, \dots$ Show that \[ \sum_{n=0}^\infty \frac{f_n(1)}{n!} = e^e. \] | Since $f_0(x) = \sum_{k=0}^\infty \frac{x^k}{k!}$, one easily shows by mathematical induction that \[ f_n(x) = \sum_{k=0}^\infty \frac{k^n x^k}{k!}. \] Then, since all terms are positive, one has \[ \sum_{n=0}^\infty \frac{f_n(1)}{n!} = \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{k^n}{k! n!} = \sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^\infty \frac{k^n}{n!} = \sum_{k=0}^\infty \frac{e^k}{k!} = e^e. \] | 0 |
1976 | 1976_A3 | Find all integral solutions of the equation \[|p^r - q^s| = 1,\] where $p$ and $q$ are prime numbers and $r$ and $s$ are positive integers larger than unity. Prove that there are no other solutions. Return the sum of all quadruples (p, q, r, s) by adding corresponding p, q, r and s values across all solutions and giving the final summed quadruple. | We show that the only solutions are given by $3^2 - 2^3 = 1$, i.e., $(p, r, q, s) = (3, 2, 2, 3)$ or $(2, 3, 3, 2)$.\
Clearly either $p$ or $q$ is $2$. Suppose $q = 2$. Then $p$ is an odd prime with $p^r \pm 1 = 2^s$. If $r$ is odd, $(p^r \pm 1)/(p \pm 1)$ is the odd integer $p^{r-1} + p^{r-2} + p^{r-3} + \cdots + 1$, which is greater than $1$ since $r > 1$; this contradicts the fact that $2^s$ has no such factor.\
Now we try $r$ as an even integer $2t$. Then $p^r + 1 = 2^s$ leads to \[ 2^s = (p^t)^2 + 1 = (2n+1)^2 + 1 = 4n^2 + 4n + 2, \] which is impossible since $4 \not| (4n^2 + 4n + 2)$.\
Also $r = 2t$ and $p^r - 1 = 2^s$ leads to \[ (p^t)^2 - 1 = (2n+1)^2 - 1 = 4n^2 + 4n = 4n(n+1) = 2^s. \] Since either $n$ or $n+1$ is odd, this is only possible for $n = 1$, $s = 3$, $p = 3$, and $r = 2$. Thus the final quadruple is \boxed{(5, 5, 5, 5)}. | numerical | putnam (modified boxing) | Number Theory | Find all integral solutions of the equation \[|p^r - q^s| = 1,\] where $p$ and $q$ are prime numbers and $r$ and $s$ are positive integers larger than unity. Prove that there are no other solutions. | We show that the only solutions are given by $3^2 - 2^3 = 1$, i.e., $\boxed{(p, r, q, s) = (3, 2, 2, 3)$ or $(2, 3, 3, 2)}$.\
Clearly either $p$ or $q$ is $2$. Suppose $q = 2$. Then $p$ is an odd prime with $p^r \pm 1 = 2^s$. If $r$ is odd, $(p^r \pm 1)/(p \pm 1)$ is the odd integer $p^{r-1} + p^{r-2} + p^{r-3} + \cdots + 1$, which is greater than $1$ since $r > 1$; this contradicts the fact that $2^s$ has no such factor.\
Now we try $r$ as an even integer $2t$. Then $p^r + 1 = 2^s$ leads to \[ 2^s = (p^t)^2 + 1 = (2n+1)^2 + 1 = 4n^2 + 4n + 2, \] which is impossible since $4 \not| (4n^2 + 4n + 2)$.\
Also $r = 2t$ and $p^r - 1 = 2^s$ leads to \[ (p^t)^2 - 1 = (2n+1)^2 - 1 = 4n^2 + 4n = 4n(n+1) = 2^s. \] Since either $n$ or $n+1$ is odd, this is only possible for $n = 1$, $s = 3$, $p = 3$, and $r = 2$. | 0 |
1976 | 1976_A4 | Let $r$ be a root of $P(x) = x^3 + ax^2 + bx - 1 = 0$ and $r + 1$ be a root of $y^3 + cy^2 + dy + 1 = 0$, where $a, b, c, d$ are integers. Also let $P(x)$ be irreducible over the rational numbers. Find all root $s$ of $P(x) = 0$ as a function of $r$ which does not explicitly involve $a, b, c, or d$ and sum them together. | We show that one answer is $s = -1/(r+1)$ and another answer is $s = -(r+1)/r = -1-(1/r)$. Since $P(x)$ is irreducible, so is $M(x) = P(x-1)$. Hence $M(x)$ is the only monic cubic over the rationals with $r+1$ as a zero, i.e., $M(x) = x^3 + cx^2 + dx + 1$. If the zeros of $P$ are $r, s, t$, and the zeros of $M$ are $r+1, s+1, t+1$. Now the coefficients $-1$ and $1$ of $x^0$ in $P$ and $M$, respectively, tell us that $rst = 1$ and $(r+1)(s+1)(t+1) = -1$. Then \[ st = \frac{1}{r}, \quad s + t = \frac{-r^2 + 3r + 1}{r(r+1)} \] Hence $s$ is either root of \[ x^2 + \frac{r^2 + 3r + 1}{r(r+1)}x + \frac{1}{r} = 0. \] Using the quadratic formula, one finds that $s$ is $-1/(r+1)$ or $-(r+1)/r$ with the final sum being \boxed{(-r^2 - 3r - 1)/(r(r+1))}. | algebraic | putnam (modified boxing) | Algebra | Let $r$ be a root of $P(x) = x^3 + ax^2 + bx - 1 = 0$ and $r + 1$ be a root of $y^3 + cy^2 + dy + 1 = 0$, where $a, b, c, d$ are integers. Also let $P(x)$ be irreducible over the rational numbers. Express another root $s$ of $P(x) = 0$ as a function of $r$ which does not explicitly involve $a, b, c, or d$. | We show that one answer is $s = -1/(r+1)$ and another answer is $s = -(r+1)/r = -1-(1/r)$. Since $P(x)$ is irreducible, so is $M(x) = P(x-1)$. Hence $M(x)$ is the only monic cubic over the rationals with $r+1$ as a zero, i.e., $M(x) = x^3 + cx^2 + dx + 1$. If the zeros of $P$ are $r, s, t$, and the zeros of $M$ are $r+1, s+1, t+1$. Now the coefficients $-1$ and $1$ of $x^0$ in $P$ and $M$, respectively, tell us that $rst = 1$ and $(r+1)(s+1)(t+1) = -1$. Then \[ st = \frac{1}{r}, \quad s + t = \frac{-r^2 + 3r + 1}{r(r+1)} \] Hence $s$ is either root of \[ x^2 + \frac{r^2 + 3r + 1}{r(r+1)}x + \frac{1}{r} = 0. \] Using the quadratic formula, one finds that $s$ is \boxed{$-1/(r+1)$ or $-(r+1)/r$}. | 0 |
1976 | 1976_A5 | In the $(x,y)$-plane, if $R$ is the set of points inside and on a convex polygon, let $D(x,y)$ be the distance from $(x,y)$ to the nearest point of $R$. (a) Show that there exist constants $a$, $b$, and $c$, independent of $R$, such that
\[
\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-D(x,y)} dx dy = a + bL + cA,
\]
where $L$ is the perimeter of $R$ and $A$ is the area of $R$. (b) Find the sum of the values of $a$, $b$, and $c$. | It is shown below that $a = 2\pi$, $b = 1$, and $c = 1$. We use $I[S]$ to denote the integral of $e^{-D(x,y)}$ over a region $S$. Since $D(x,y) = 0$ on $R$, $I[R] = A$. Now let $\sigma$ be a side of $R$, and $S(\sigma)$ be the half strip consisting of the points of the plane having a point on $\sigma$ as the nearest point of $R$. Changing to $(u,v)$-coordinates with $u$ measured parallel to $\sigma$ and $v$ measured perpendicular to $\sigma$, one finds that $I[S(\sigma)] = \int_0^\infty \int_0^\infty e^{-v} dv du = s$. The sum $\Sigma_1$ of these integrals for all the sides of $R$ is $L$.
If $v$ is a vertex of $R$, the points with $v$ as the nearest point of $R$ lie in the inside $T(v)$ of an angle bounded by the rays from $v$ perpendicular to the edges meeting at $v$; let $\alpha = \alpha(v)$ be the measure of this angle. Using polar coordinates, one has
\[
I[T(v)] = \int_0^\alpha \int_0^\infty r e^{-r} dr d\theta = \alpha.
\]
The sum $\Sigma_2$ of the $I[T(v)]$ for all vertices $v$ of $R$ is $2\pi$. Now the original double integral equals $\Sigma_2 + \Sigma_1 + A = 2\pi + L + A$. Hence $a = 2\pi$, $b = 1$, and $c = 1$ with the final answer being \boxed{2\pi + 2}. | numerical | putnam (modified boxing) | Geometry Calculus | In the $(x,y)$-plane, if $R$ is the set of points inside and on a convex polygon, let $D(x,y)$ be the distance from $(x,y)$ to the nearest point of $R$. (a) Show that there exist constants $a$, $b$, and $c$, independent of $R$, such that
\[
\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-D(x,y)} dx dy = a + bL + cA,
\]
where $L$ is the perimeter of $R$ and $A$ is the area of $R$. (b) Find the values of $a$, $b$, and $c$. | It is shown below that $a = 2\pi$, $b = 1$, and $c = 1$. We use $I[S]$ to denote the integral of $e^{-D(x,y)}$ over a region $S$. Since $D(x,y) = 0$ on $R$, $I[R] = A$. Now let $\sigma$ be a side of $R$, and $S(\sigma)$ be the half strip consisting of the points of the plane having a point on $\sigma$ as the nearest point of $R$. Changing to $(u,v)$-coordinates with $u$ measured parallel to $\sigma$ and $v$ measured perpendicular to $\sigma$, one finds that $I[S(\sigma)] = \int_0^\infty \int_0^\infty e^{-v} dv du = s$. The sum $\Sigma_1$ of these integrals for all the sides of $R$ is $L$.
If $v$ is a vertex of $R$, the points with $v$ as the nearest point of $R$ lie in the inside $T(v)$ of an angle bounded by the rays from $v$ perpendicular to the edges meeting at $v$; let $\alpha = \alpha(v)$ be the measure of this angle. Using polar coordinates, one has
\[
I[T(v)] = \int_0^\alpha \int_0^\infty r e^{-r} dr d\theta = \alpha.
\]
The sum $\Sigma_2$ of the $I[T(v)]$ for all vertices $v$ of $R$ is $2\pi$. Now the original double integral equals $\Sigma_2 + \Sigma_1 + A = 2\pi + L + A$. Hence \boxed{$a = 2\pi$, $b = 1$, and $c = 1}$. | 0 |
1976 | 1976_B1 | \text{Evaluate } \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \left\lfloor \frac{2n}{k} \right\rfloor - 2 \left\lfloor \frac{n}{k} \right\rfloor \right) \text{ and express your answer in the form } \log a - b, \text{ with } a \text{ and } b \text{ positive integers.} \\ \text{Here } \lfloor x \rfloor \text{ is defined to be the integer such that } \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 \text{ and } \log x \text{ is the logarithm of } x \text{ to base } e. | It is shown below that $a = 4$ and $b = 1$. Let $f(x) = \lfloor 2/x \rfloor - 2\lfloor 1/x \rfloor$. Then the desired limit $L$ equals $\int_0^1 f(x)dx$. For $n = 1, 2, \dots$, $f(x) = 0$ on $2/(2n + 1) < x \leq 1/n$ and $f(x) = 1$ on $1/(n + 1) < x \leq 2/(2n + 1)$. Hence \\ \[ L = \left( \frac{2}{3} - \frac{2}{4} \right) + \left( \frac{2}{5} - \frac{2}{6} \right) + \cdots = -1 + 2 \left( 1 - \frac{1}{2} + \frac{1}{3} - \cdots \right) \] \\ \[ = -1 + 2 \int_0^1 \frac{dx}{1+x} = -1 + 2 \ln 2 = \boxed{\ln 4 - 1}. \] | algebraic | putnam | Analysis | \text{Evaluate } \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \left\lfloor \frac{2n}{k} \right\rfloor - 2 \left\lfloor \frac{n}{k} \right\rfloor \right) \text{ and express your answer in the form } \log a - b, \text{ with } a \text{ and } b \text{ positive integers.} \\ \text{Here } \lfloor x \rfloor \text{ is defined to be the integer such that } \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 \text{ and } \log x \text{ is the logarithm of } x \text{ to base } e. | It is shown below that $a = 4$ and $b = 1$. Let $f(x) = \lfloor 2/x \rfloor - 2\lfloor 1/x \rfloor$. Then the desired limit $L$ equals $\int_0^1 f(x)dx$. For $n = 1, 2, \dots$, $f(x) = 0$ on $2/(2n + 1) < x \leq 1/n$ and $f(x) = 1$ on $1/(n + 1) < x \leq 2/(2n + 1)$. Hence \\ \[ L = \left( \frac{2}{3} - \frac{2}{4} \right) + \left( \frac{2}{5} - \frac{2}{6} \right) + \cdots = -1 + 2 \left( 1 - \frac{1}{2} + \frac{1}{3} - \cdots \right) \] \\ \[ = -1 + 2 \int_0^1 \frac{dx}{1+x} = -1 + 2 \ln 2 = \boxed{\ln 4 - 1}. \] | 0 |
1976 | 1976_B2 | Suppose that $G$ is a group generated by elements $A$ and $B$, that is, every element of $G$ can be written as a finite "word" $A^{n_1}B^{n_2}A^{n_3}\cdots B^{n_k}$, where $n_1, \ldots, n_k$ are any integers, and $A^0 = B^0 = 1$ as usual. Also, suppose that $A^4 = B^7 = ABA^{-1}B = 1$, $A^2 \neq 1$, and $B \neq 1$.\n\n(a) How many elements of $G$ are of the form $C^2$ with $C$ in $G$?\n\n(b) Write each such square as a word in $A$ and $B$ As the final answer return an expression adding the number from (a) and all the squares in terms of $A$ and $B$ from (b). | The answers are (a) $8$; (b) $1, A^2, B, B^2, B^3, B^4, B^5, B^6$. Since $B = (B^7)^2, B^3 = (B^5)^2, B^5 = (B^6)^2$, the elements in the answer to (b) are all squares in $G$. They are distinct since $B$ has order $7$ and $A$ has order $4$. To show that there are no other squares, we first note that $ABA^{-1}B = 1$ implies $AB = B^{-1}A$.\n\nThen\n\n\[ AB^2 = (B^{-1}A)B = B^{-1}(AB) = B^{-1}(B^{-1}A) = B^{-2}A. \]\n\nSimilarly $AB^n = B^{-n}A$ for the other $n$'s in $\{0,1,\ldots,6\}$ and so for all integers $n$. Thus the final answer is \boxed{9 + A^2 + B + B^2 + B^3 + B^4 + B^5 + B^6}. | algebraic | putnam (modified boxing) | Algebra Number Theory | Suppose that $G$ is a group generated by elements $A$ and $B$, that is, every element of $G$ can be written as a finite "word" $A^{n_1}B^{n_2}A^{n_3}\cdots B^{n_k}$, where $n_1, \ldots, n_k$ are any integers, and $A^0 = B^0 = 1$ as usual. Also, suppose that $A^4 = B^7 = ABA^{-1}B = 1$, $A^2 \neq 1$, and $B \neq 1$.\n\n(a) How many elements of $G$ are of the form $C^2$ with $C$ in $G$?\n\n(b) Write each such square as a word in $A$ and $B$. | The answers are \boxed{(a) $8$; (b) $1, A^2, B, B^2, B^3, B^4, B^5, B^6$}. Since $B = (B^7)^2, B^3 = (B^5)^2, B^5 = (B^6)^2$, the elements in the answer to (b) are all squares in $G$. They are distinct since $B$ has order $7$ and $A$ has order $4$. To show that there are no other squares, we first note that $ABA^{-1}B = 1$ implies $AB = B^{-1}A$.\n\nThen\n\n\[ AB^2 = (B^{-1}A)B = B^{-1}(AB) = B^{-1}(B^{-1}A) = B^{-2}A. \]\n\nSimilarly $AB^n = B^{-n}A$ for the other $n$'s in $\{0,1,\ldots,6\}$ and so for all integers $n$. | 0 |
1976 | 1976_B5 | \[ \sum_{k=1}^n (-1)^k \binom{n}{k} (x-k)^n. \] | The sum is \( n! \) since it is an \( n \)-th difference of a monic polynomial, \( x^n \), of degree \( n \) The final answer is \boxed{n!}. | numerical | putnam | Algebra | \[ \sum_{k=1}^n (-1)^k \binom{n}{k} (x-k)^n. \] | The sum is \( \boxed{n!} \) since it is an \( n \)-th difference of a monic polynomial, \( x^n \), of degree \( n \). | 0 |
1977 | 1977_A1 | Consider all lines which meet the graph of \[ y = 2x^4 + 7x^3 + 3x - 5 \] in four distinct points, say $(x_i, y_i)$, $i=1,2,3,4$. Show that \[ \frac{x_1 + x_2 + x_3 + x_4}{4} \] is independent of the line and find its value. | A line meeting the graph in four points has an equation $y = mx + b$. Then the $x_i$ are the roots of \[ 2x^4 + 7x^3 + (3-m)x - (5+b) = 0, \] their sum is $-\frac{7}{2}$, and their arithmetic mean $(\Sigma x_i)/4$ is $\boxed{-\frac{7}{8}}$, which is independent of the line. | numerical | putnam | Algebra | Consider all lines which meet the graph of \[ y = 2x^4 + 7x^3 + 3x - 5 \] in four distinct points, say $(x_i, y_i)$, $i=1,2,3,4$. Show that \[ \frac{x_1 + x_2 + x_3 + x_4}{4} \] is independent of the line and find its value. | A line meeting the graph in four points has an equation $y = mx + b$. Then the $x_i$ are the roots of \[ 2x^4 + 7x^3 + (3-m)x - (5+b) = 0, \] their sum is $-\frac{7}{2}$, and their arithmetic mean $(\Sigma x_i)/4$ is $\boxed{-\frac{7}{8}}$, which is independent of the line. | 0 |
1977 | 1977_A3 | Let $u, f, \text{and } g$ be functions, defined for all real numbers $x$, such that \[ \frac{u(x+1) + u(x-1)}{2} = f(x) \quad \text{and} \quad \frac{u(x+4) + u(x-4)}{2} = g(x). \] Determine $u(x)$ in terms of $f$ and $g$. | We show that there are an infinite number of expressions for $u(x)$ in terms of $f$ and $g$; some of the simpler ones are: \[ u(x) = g(x) - f(x+3) + f(x+1) + f(x-1) - f(x-3), \] \[ = -g(x+2) + f(x+5) - f(x+3) + f(x+1) + f(x-1), \] \[ = g(x+4) - f(x+7) + f(x+5) - f(x+3) + f(x+1). \] Let $E$ be the shift operator on functions $A$ defined by $EA(x) = A(x+1)$. Then $(E+E^{-1})u(x) = 2f(x)$ and $(E^4+E^{-4})u(x) = 2g(x)$ are given. Thus \[ (E^2+1)u(x) = 2Ef(x) \] and \[ (E^8+1)u(x) = 2E^4g(x). \] Motivated by the fact that $E^2+1$ and $E^8+1$ are relatively prime polynomials in $E$, one finds that \[ 1 = \frac{1}{2}(E^8+1) - \frac{1}{2}(E^6-E^4+E^2-1)(E^2+1), \] \[ u(x) = \frac{1}{2}(E^8+1)u(x) - \frac{1}{2}(E^6-E^4+E^2-1)(E^2+1)u(x), \] \[ u(x) = E^4g(x) - (E^6-E^4+E^2-1)Ef(x), \] \[ u(x) = E^4g(x) + (-E^7+E^5-E^3+E)f(x), \] \[ u(x) = \boxed{g(x+4) - f(x+7) + f(x+5) - f(x+3) + f(x+1)}. \] | algebraic | putnam | Algebra Analysis | Let $u, f, \text{and } g$ be functions, defined for all real numbers $x$, such that \[ \frac{u(x+1) + u(x-1)}{2} = f(x) \quad \text{and} \quad \frac{u(x+4) + u(x-4)}{2} = g(x). \] Determine $u(x)$ in terms of $f$ and $g$. | We show that there are an infinite number of expressions for $u(x)$ in terms of $f$ and $g$; some of the simpler ones are: \[ u(x) = g(x) - f(x+3) + f(x+1) + f(x-1) - f(x-3), \] \[ = -g(x+2) + f(x+5) - f(x+3) + f(x+1) + f(x-1), \] \[ = g(x+4) - f(x+7) + f(x+5) - f(x+3) + f(x+1). \] Let $E$ be the shift operator on functions $A$ defined by $EA(x) = A(x+1)$. Then $(E+E^{-1})u(x) = 2f(x)$ and $(E^4+E^{-4})u(x) = 2g(x)$ are given. Thus \[ (E^2+1)u(x) = 2Ef(x) \] and \[ (E^8+1)u(x) = 2E^4g(x). \] Motivated by the fact that $E^2+1$ and $E^8+1$ are relatively prime polynomials in $E$, one finds that \[ 1 = \frac{1}{2}(E^8+1) - \frac{1}{2}(E^6-E^4+E^2-1)(E^2+1), \] \[ u(x) = \frac{1}{2}(E^8+1)u(x) - \frac{1}{2}(E^6-E^4+E^2-1)(E^2+1)u(x), \] \[ u(x) = E^4g(x) - (E^6-E^4+E^2-1)Ef(x), \] \[ u(x) = E^4g(x) + (-E^7+E^5-E^3+E)f(x), \] \[ u(x) = \boxed{g(x+4) - f(x+7) + f(x+5) - f(x+3) + f(x+1)}. \] | 0 |
1977 | 1977_A4 | For $0 < x < 1$, express
\[ \sum_{n=0}^\infty \frac{x^{2^n}}{1 - x^{2^{n+1}}} \]
as a rational function of $x$. | \[ \sum_{n=0}^N \frac{x^{2^n}}{1 - x^{2^{n+1}}} = \sum_{n=0}^N \left( \frac{1}{1 - x^{2^n}} - \frac{1}{1 - x^{2^{n+1}}} \right) = \frac{1}{1 - x} - \frac{1}{1 - x^{2^{N+1}}} \to \frac{1}{1 - x} - 1 = \boxed{\frac{x}{1 - x}} \text{ as } N \to \infty, \]
since $|x| < 1$. | algebraic | putnam | Analysis Algebra | For $0 < x < 1$, express
\[ \sum_{n=0}^\infty \frac{x^{2^n}}{1 - x^{2^{n+1}}} \]
as a rational function of $x$. | \[ \sum_{n=0}^N \frac{x^{2^n}}{1 - x^{2^{n+1}}} = \sum_{n=0}^N \left( \frac{1}{1 - x^{2^n}} - \frac{1}{1 - x^{2^{n+1}}} \right) = \frac{1}{1 - x} - \frac{1}{1 - x^{2^{N+1}}} \to \frac{1}{1 - x} - 1 = \boxed{\frac{x}{1 - x}} \text{ as } N \to \infty, \]
since $|x| < 1$. | 0 |
1977 | 1977_B1 | Evaluate the infinite product \[ \prod_{n=2}^{\infty} \frac{n^3 - 1}{n^3 + 1}. \] | \[ \prod_{n=2}^{\infty} \frac{n^3 - 1}{n^3 + 1} = \lim_{k \to \infty} \left[ \frac{2^3 - 1}{2^3 + 1} \cdot \frac{3^3 - 1}{3^3 + 1} \cdot \ldots \cdot \frac{k^3 - 1}{k^3 + 1} \right] \\ = \lim_{k \to \infty} \left[ \frac{1 \cdot 7 \cdot 2 \cdot 13 \cdot 3 \cdot 21 \cdots (k-1)(k^2 + k + 1)}{3 \cdot 3 \cdot 4 \cdot 7 \cdot 5 \cdot 13 \cdots (k+1)(k^2 - k + 1)} \right] \\ = \lim_{k \to \infty} \left[ \frac{2}{3} \cdot \frac{k^2 + k + 1}{k(k+1)} \right] = \boxed{\frac{2}{3}}. \] | numerical | putnam | Calculus Algebra | Evaluate the infinite product \[ \prod_{n=2}^{\infty} \frac{n^3 - 1}{n^3 + 1}. \] | \[ \prod_{n=2}^{\infty} \frac{n^3 - 1}{n^3 + 1} = \lim_{k \to \infty} \left[ \frac{2^3 - 1}{2^3 + 1} \cdot \frac{3^3 - 1}{3^3 + 1} \cdot \ldots \cdot \frac{k^3 - 1}{k^3 + 1} \right] \\ = \lim_{k \to \infty} \left[ \frac{1 \cdot 7 \cdot 2 \cdot 13 \cdot 3 \cdot 21 \cdots (k-1)(k^2 + k + 1)}{3 \cdot 3 \cdot 4 \cdot 7 \cdot 5 \cdot 13 \cdots (k+1)(k^2 - k + 1)} \right] \\ = \lim_{k \to \infty} \left[ \frac{2}{3} \cdot \frac{k^2 + k + 1}{k(k+1)} \right] = \boxed{\frac{2}{3}}. \] | 0 |
1977 | 1977_B6 | Let $H$ be a subgroup with $h$ elements in a group $G$. Suppose that $G$ has an element $a$ such that for all $x$ in $H$, $(xa)^3 = 1$, the identity. In $G$, let $P$ be the subset of all products $x_1ax_2a \cdots x_na$, with $n$ a positive integer and the $x_i$ in $H$. Show that $P$ is a finite set and find the maximum number of elements in terms of $h$s. | Clearly $1 \in H$. Also $x \in H$ implies $x^{-1} \in H$. Then the hypothesis implies that $a^{-1} = a^2$ and that $xaxa = 1 = x^{-1}a^{-1}ax^{-1}a$ when $x \in H$. Thus one easily shows that \begin{enumerate} \item[(i)] $axa = x^{-1}a^2x^{-1},$ \item[(ii)] $a^2xa^2 = x^{-1}a^{-1}x^{-1}$. \end{enumerate} Let \[ A = \{xay : x,y \in H \}, \quad B = \{xa^2y : x,y \in H \}, \quad C = \{xa^2ya : x,y \in H \}, \quad \text{and} \quad Q = A \cup B \cup C. \] Each of $A, B, C$ has at most $h^2$ elements; hence $Q$ has at most $3h^2$ elements. Thus it suffices to prove that $x_1ax_2a \cdots x_na \in Q$ when each $x_i \in H$. We do this by induction on $n$.\nFor $n = 1$, one sees that $x_1a = x_1a \cdot 1 \in A \subset Q$. Now let $x_1,x_2,\ldots,x_k,x_{k+1} \in H$; $x_1ax_2a \cdots x_ka = q, x_{k+1}a = z, \text{and } qza = p$. Inductively, we assume $q \in Q$ and seek to show $p \in Q$. The assumption implies that $q$ is in $A, B, \text{or } C$. If $q = xay \in A$, then \[ p = (xay)za = xa(yz)a = x(yz)^{-1}a^2(yz)^{-1} \in B \subset Q, \] using (i). If $q = xa^2y \in B$, then \[ p = xa^2yza \in C \subset Q. \] If $q = xa^2ya \in C$, then \[ p = xa^2y(aza) = xa^2y(x^{-1}a^{-1}x^{-1})z^{-1} = x[a^2(yz^{-1})a^2]z^{-1} = x(yz^{-1})^{-1}a(yz^{-1})^{-1}z^{-1} \in A \subset Q, \] using (i) and (ii). Thus the maximum number of elements becomes $\boxed{3h^2}$. | algebraic | putnam (modified boxing) | Algebra | Let $H$ be a subgroup with $h$ elements in a group $G$. Suppose that $G$ has an element $a$ such that for all $x$ in $H$, $(xa)^3 = 1$, the identity. In $G$, let $P$ be the subset of all products $x_1ax_2a \cdots x_na$, with $n$ a positive integer and the $x_i$ in $H$.\begin{enumerate} \item[(a)] Show that $P$ is a finite set. \item[(b)] Show that, in fact, $P$ has no more than $3h^2$ elements. \end{enumerate} | Clearly $1 \in H$. Also $x \in H$ implies $x^{-1} \in H$. Then the hypothesis implies that $a^{-1} = a^2$ and that $xaxa = 1 = x^{-1}a^{-1}ax^{-1}a$ when $x \in H$. Thus one easily shows that \begin{enumerate} \item[(i)] $axa = x^{-1}a^2x^{-1},$ \item[(ii)] $a^2xa^2 = x^{-1}a^{-1}x^{-1}$. \end{enumerate} Let \[ A = \{xay : x,y \in H \}, \quad B = \{xa^2y : x,y \in H \}, \quad C = \{xa^2ya : x,y \in H \}, \quad \text{and} \quad Q = A \cup B \cup C. \] Each of $A, B, C$ has at most $h^2$ elements; hence $Q$ has at most $3h^2$ elements. Thus it suffices to prove that $x_1ax_2a \cdots x_na \in Q$ when each $x_i \in H$. We do this by induction on $n$.\nFor $n = 1$, one sees that $x_1a = x_1a \cdot 1 \in A \subset Q$. Now let $x_1,x_2,\ldots,x_k,x_{k+1} \in H$; $x_1ax_2a \cdots x_ka = q, x_{k+1}a = z, \text{and } qza = p$. Inductively, we assume $q \in Q$ and seek to show $p \in Q$. The assumption implies that $q$ is in $A, B, \text{or } C$. If $q = xay \in A$, then \[ p = (xay)za = xa(yz)a = x(yz)^{-1}a^2(yz)^{-1} \in B \subset Q, \] using (i). If $q = xa^2y \in B$, then \[ p = xa^2yza \in C \subset Q. \] If $q = xa^2ya \in C$, then \[ p = xa^2y(aza) = xa^2y(x^{-1}a^{-1}x^{-1})z^{-1} = x[a^2(yz^{-1})a^2]z^{-1} = x(yz^{-1})^{-1}a(yz^{-1})^{-1}z^{-1} \in A \subset Q, \] using (i) and (ii). | 0 |
1978 | 1978_A3 | Let $p(x) = 2 + 4x + 3x^2 + 5x^3 + 3x^4 + 4x^5 + 2x^6$. For $k$ with $0 < k < 5$, define \[ I_k = \int_0^\infty \frac{x^k}{p(x)} dx. \] For which $k$ is $I_k$ smallest? | Since the integral converges for $-1 < k < 5$, one can consider $I_k$ to be defined on this open interval. Letting $x = 1/t$, one finds that \[ I_k = \int_0^\infty \frac{t^{-k}}{t^{-6}p(t)} \left( -\frac{dt}{t^2} \right) = \int_0^\infty \frac{t^{4-k} dt}{p(t)} = I_{4-k}. \] Then \[ I_k = (I_k + I_{4-k})/2 = \int_0^\infty \frac{[(x^k + x^{4-k})/2]}{p(x)} dx > \int_0^\infty \frac{x^2 dx}{p(x)} = I_2, \] since $(x^k + x^{4-k})/2 \geq \sqrt{x^k \cdot x^{4-k}} = x^2$ by the Arithmetic Mean–Geometric Mean Inequality. Thus $I_k$ is smallest for $k = \boxed{2}$. | numerical | putnam | Calculus Analysis | Let $p(x) = 2 + 4x + 3x^2 + 5x^3 + 3x^4 + 4x^5 + 2x^6$. For $k$ with $0 < k < 5$, define \[ I_k = \int_0^\infty \frac{x^k}{p(x)} dx. \] For which $k$ is $I_k$ smallest? | Since the integral converges for $-1 < k < 5$, one can consider $I_k$ to be defined on this open interval. Letting $x = 1/t$, one finds that \[ I_k = \int_0^\infty \frac{t^{-k}}{t^{-6}p(t)} \left( -\frac{dt}{t^2} \right) = \int_0^\infty \frac{t^{4-k} dt}{p(t)} = I_{4-k}. \] Then \[ I_k = (I_k + I_{4-k})/2 = \int_0^\infty \frac{[(x^k + x^{4-k})/2]}{p(x)} dx > \int_0^\infty \frac{x^2 dx}{p(x)} = I_2, \] since $(x^k + x^{4-k})/2 \geq \sqrt{x^k \cdot x^{4-k}} = x^2$ by the Arithmetic Mean–Geometric Mean Inequality. Thus $I_k$ is smallest for $k = \boxed{2}$. | 0 |
1978 | 1978_A6 | Let $n$ distinct points in the plane be given. Find the upper bound of the number of pairs of them that are unit distance apart. | Let $f(n)$ be the maximum number of pairs with unit distance apart from a set of $n$ points in a plane. Then $f(1) = 0, f(2) = 1, f(3) = 3, \text{ and } f(4) = 5.\\ \text{Suppose we have an array of } n \geq 2 \text{ points which realizes } f(n) \text{ pairs unit distance apart. Suppose}\\ \text{each point has at least } k \text{ points (in the set) unit distance away and that one of them, namely } x_0,\\ \text{has exactly } k \text{ points unit distance away. Let these points be } x_1, \dots, x_k.\\ \text{For } i = 0, 1, \dots, k, \text{ let } C_i \text{ be the circle with radius 1 and center } x_i. \text{ For } i > 0, C_i \text{ goes through } x_0,\\ \text{and at most two other } x_j. \text{ Therefore, there are at least } k - 3 \text{ points of the array other than }\\ x_0, \dots, x_k \text{ on } C_i. \text{ Any one of these } k - 3 \text{ points can appear on at most one other } C_j \text{ since there are}\\ \text{only two unit circles through } x_0 \text{ and another point. Hence}\\ n \geq 1 + k + \frac{k(k-3)}{2} = 1 + \frac{k(k-1)}{2}.\\ \text{Thus } k \text{ is such that the triangular number } \frac{k(k-1)}{2} \text{ is less than } n; \text{ we let the largest such integer be } k_n.\\ \text{The array with } x_0 \text{ removed has at most } f(n-1) \text{ pairs unit distance apart; therefore } f(n) \leq f(n-1) + k_n. \\ \text{By repeating this argument we find that } f(n) \leq k_2 + k_3 + \cdots + k_n.\\ \text{The definition of } k_n \text{ implies that } k_n = t \text{ for } \binom{t}{2} < n \leq \binom{t+1}{2}. \text{ From this it follows that } k_n \leq 1 + \sqrt{2(n-1)}.\\ \text{Hence}\\ f(n) \leq k_2 + k_3 + \cdots + k_n\\ \leq n-1 + \sqrt{2}(\sqrt{1} + \sqrt{2} + \cdots + \sqrt{n-1})\\ \leq n-1 + \sqrt{2} \int_{1}^{n} \sqrt{x} \, dx\\ \leq n - 1 + 2\sqrt{2} \cdot \frac{n^{3/2} - 1}{3}\\ \leq n + .95n^{3/2} < \boxed{2n^{3/2}}. | algebraic | putnam (modified boxing) | Geometry Analysis | Let $n$ distinct points in the plane be given. Prove that fewer than $2n^{3/2}$ pairs of them are unit distance apart. | Let $f(n)$ be the maximum number of pairs with unit distance apart from a set of $n$ points in a plane. Then $f(1) = 0, f(2) = 1, f(3) = 3, \text{ and } f(4) = 5.\\ \text{Suppose we have an array of } n \geq 2 \text{ points which realizes } f(n) \text{ pairs unit distance apart. Suppose}\\ \text{each point has at least } k \text{ points (in the set) unit distance away and that one of them, namely } x_0,\\ \text{has exactly } k \text{ points unit distance away. Let these points be } x_1, \dots, x_k.\\ \text{For } i = 0, 1, \dots, k, \text{ let } C_i \text{ be the circle with radius 1 and center } x_i. \text{ For } i > 0, C_i \text{ goes through } x_0,\\ \text{and at most two other } x_j. \text{ Therefore, there are at least } k - 3 \text{ points of the array other than }\\ x_0, \dots, x_k \text{ on } C_i. \text{ Any one of these } k - 3 \text{ points can appear on at most one other } C_j \text{ since there are}\\ \text{only two unit circles through } x_0 \text{ and another point. Hence}\\ n \geq 1 + k + \frac{k(k-3)}{2} = 1 + \frac{k(k-1)}{2}.\\ \text{Thus } k \text{ is such that the triangular number } \frac{k(k-1)}{2} \text{ is less than } n; \text{ we let the largest such integer be } k_n.\\ \text{The array with } x_0 \text{ removed has at most } f(n-1) \text{ pairs unit distance apart; therefore } f(n) \leq f(n-1) + k_n. \\ \text{By repeating this argument we find that } f(n) \leq k_2 + k_3 + \cdots + k_n.\\ \text{The definition of } k_n \text{ implies that } k_n = t \text{ for } \binom{t}{2} < n \leq \binom{t+1}{2}. \text{ From this it follows that } k_n \leq 1 + \sqrt{2(n-1)}.\\ \text{Hence}\\ f(n) \leq k_2 + k_3 + \cdots + k_n\\ \leq n-1 + \sqrt{2}(\sqrt{1} + \sqrt{2} + \cdots + \sqrt{n-1})\\ \leq n-1 + \sqrt{2} \int_{1}^{n} \sqrt{x} \, dx\\ \leq n - 1 + 2\sqrt{2} \cdot \frac{n^{3/2} - 1}{3}\\ \leq n + .95n^{3/2} < 2n^{3/2}. | 0 |
1978 | 1978_B1 | Find the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form $r + s \sqrt{t}$ with $r, s,$ and $t$ positive integers. | The area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure $\frac{3\pi}{4}$ and one can augment the octagon into a square with sides of length $3 + 2\sqrt{2}$ by properly placing a $\sqrt{2}, \sqrt{2}, 2$ isosceles right triangle on each of the sides of length 2. Hence the desired area is \[ (3 + 2\sqrt{2})^2 - (4\sqrt{2} \cdot \sqrt{2}/2) = 13 + 12\sqrt{2}. \]
A second solution follows. Let $r$ be the radius of the circle and let $\alpha$ and $\beta$ be half of the central angles for the chords of lengths 3 and 2, respectively. Then $8\alpha + 8\beta = 2\pi$ and so $\beta = \frac{\pi}{4} - \alpha$. Also \[ \frac{3}{2r} = \sin \alpha, \quad \frac{1}{r} = \sin \beta = \sin \left( \frac{\pi}{4} - \alpha \right) = \frac{\cos \alpha - \sin \alpha}{\sqrt{2}}, \quad \frac{2}{3} = \frac{2r}{3} \cdot \frac{1}{r} = \frac{\cos \alpha - \sin \alpha}{\sqrt{2} \sin \alpha}. \] Now $\cot \alpha = (3 + 2\sqrt{2})/3 = [(3 + 2\sqrt{2})/2]/(3/2)$ and hence the distance from the center of the circle to a chord of length 3 is $h_3 = (3 + 2\sqrt{2})/2$. Similarly the distance to a chord of length 2 is $h_2 = (2 + 3\sqrt{2})/2$. Finally, the desired area is \[ 4(3h_3 + 2h_2)/2 = (9 + 6\sqrt{2}) + (4 + 6\sqrt{2}) = \boxed{13 + 12\sqrt{2}}. \] | numerical | putnam | Geometry | Find the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form $r + s \sqrt{t}$ with $r, s,$ and $t$ positive integers. | The area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure $\frac{3\pi}{4}$ and one can augment the octagon into a square with sides of length $3 + 2\sqrt{2}$ by properly placing a $\sqrt{2}, \sqrt{2}, 2$ isosceles right triangle on each of the sides of length 2. Hence the desired area is \[ (3 + 2\sqrt{2})^2 - (4\sqrt{2} \cdot \sqrt{2}/2) = 13 + 12\sqrt{2}. \]
A second solution follows. Let $r$ be the radius of the circle and let $\alpha$ and $\beta$ be half of the central angles for the chords of lengths 3 and 2, respectively. Then $8\alpha + 8\beta = 2\pi$ and so $\beta = \frac{\pi}{4} - \alpha$. Also \[ \frac{3}{2r} = \sin \alpha, \quad \frac{1}{r} = \sin \beta = \sin \left( \frac{\pi}{4} - \alpha \right) = \frac{\cos \alpha - \sin \alpha}{\sqrt{2}}, \quad \frac{2}{3} = \frac{2r}{3} \cdot \frac{1}{r} = \frac{\cos \alpha - \sin \alpha}{\sqrt{2} \sin \alpha}. \] Now $\cot \alpha = (3 + 2\sqrt{2})/3 = [(3 + 2\sqrt{2})/2]/(3/2)$ and hence the distance from the center of the circle to a chord of length 3 is $h_3 = (3 + 2\sqrt{2})/2$. Similarly the distance to a chord of length 2 is $h_2 = (2 + 3\sqrt{2})/2$. Finally, the desired area is \[ 4(3h_3 + 2h_2)/2 = (9 + 6\sqrt{2}) + (4 + 6\sqrt{2}) = \boxed{13 + 12\sqrt{2}}. \] | 0 |
1978 | 1978_B2 | Express
\[ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^2 n + mn^2 + 2mn} \]
as a rational number. | Let $S$ be the desired sum. Then
\[ S = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{n+2} \left( \frac{1}{m} - \frac{1}{m+n+2} \right) \]
\[ = \sum_{n=1}^{\infty} \frac{1}{n(n+2)} \left([1 - \frac{1}{n+3}] + [\frac{1}{2} - \frac{1}{n+4}] + [\frac{1}{3} - \frac{1}{n+5}] + \cdots \right). \]
Hence
\[ 2S = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \lim_{k \to \infty} \left[1 + \frac{1}{2} + \cdots + \frac{1}{n+2} - \frac{1}{k} - \frac{1}{k+1} - \cdots - \frac{1}{k+n+1} \right]. \]
\[ = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \left(1 + \frac{1}{2} + \cdots + \frac{1}{n+2} \right). \]
\[ = \lim_{h \to \infty} \left( [(1 - \frac{1}{3})(1 + \frac{1}{2} + \frac{1}{3})] + [(\frac{1}{2} - \frac{1}{4})(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})] + \cdots \right) \]
\[ = \lim_{h \to \infty} \left( [(1)(1 + \frac{1}{2} + \frac{1}{3})] + [\frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})] + \cdots \right). \]
\[ = \frac{6+3+2}{6} + \frac{12+6+4+3}{2 \cdot 12} + \left( \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots \right) \]
\[ = \frac{11}{6} + \frac{25}{24} + \frac{1}{2} \left( \frac{1}{3} + \frac{1}{4} \right) + \frac{7}{2}. \]
\[ \text{Thus } S = \boxed{\frac{7}{4}}. \] | numerical | putnam | Analysis Number Theory | Express
\[ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^2 n + mn^2 + 2mn} \]
as a rational number. | Let $S$ be the desired sum. Then
\[ S = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{n+2} \left( \frac{1}{m} - \frac{1}{m+n+2} \right) \]
\[ = \sum_{n=1}^{\infty} \frac{1}{n(n+2)} \left([1 - \frac{1}{n+3}] + [\frac{1}{2} - \frac{1}{n+4}] + [\frac{1}{3} - \frac{1}{n+5}] + \cdots \right). \]
Hence
\[ 2S = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \lim_{k \to \infty} \left[1 + \frac{1}{2} + \cdots + \frac{1}{n+2} - \frac{1}{k} - \frac{1}{k+1} - \cdots - \frac{1}{k+n+1} \right]. \]
\[ = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \left(1 + \frac{1}{2} + \cdots + \frac{1}{n+2} \right). \]
\[ = \lim_{h \to \infty} \left( [(1 - \frac{1}{3})(1 + \frac{1}{2} + \frac{1}{3})] + [(\frac{1}{2} - \frac{1}{4})(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})] + \cdots \right) \]
\[ = \lim_{h \to \infty} \left( [(1)(1 + \frac{1}{2} + \frac{1}{3})] + [\frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})] + \cdots \right). \]
\[ = \frac{6+3+2}{6} + \frac{12+6+4+3}{2 \cdot 12} + \left( \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots \right) \]
\[ = \frac{11}{6} + \frac{25}{24} + \frac{1}{2} \left( \frac{1}{3} + \frac{1}{4} \right) + \frac{7}{2}. \]
\[ \text{Thus } S = \boxed{\frac{7}{4}}. \] | 0 |
1978 | 1978_B3 | The sequence \( \{Q_n(x)\} \) of polynomials is defined by
\[ Q_1(x) = 1 + x, \quad Q_2(x) = 1 + 2x, \]
and, for \( m > 1 \), by
\[ Q_{2m+1}(x) = Q_{2m}(x) + (m+1)xQ_{2m-1}(x), \quad Q_{2m+2}(x) = Q_{2m+1}(x) + (m+1)xQ_{2m}(x). \]
Let \( x_n \) be the largest real solution of \( Q_n(x) = 0 \). Prove that \( \{x_n\} \) is an increasing sequence and find the value of \lim_{n \to \infty} x_n. | Clearly, \(x_1 = -1, \ x_2 = -\frac{1}{2} \). An easy induction shows that each \( Q_n \) is positive for \(x \geq 0\). Hence \( x_n < 0 \), if \( Q_n \) has zeros.
Assume inductively that \( x_1 < x_2 < \cdots < x_{2m-1} < x_{2m} \). Then \( Q_{2m-1}(x) > 0 \) for \( x > x_{2m-1} \). In particular,
\[ Q_{2m+1}(x_{2m}) = Q_{2m}(x_{2m}) + (m+1)x_{2m}Q_{2m-1}(x_{2m}) < 0. \]
This implies that \( Q_{2m+1}(x) = 0 \) for some \( x > x_{2m} \), i.e., \( x_{2m+1} > x_{2m} \). Similarly, one shows that \( x_{2m+2} > x_{2m+1} \).
Let \( a = -\frac{1}{m+1} \). Using the given recursive definition of the \( Q_n(x) \), one finds that
\[ Q_{2m+2}(a) = Q_{2m+1}(a) - Q_{2m}(a) = -Q_{2m-1}(a). \]
Hence at least one of \( Q_{2m+2}(a) \) and \( Q_{2m-1}(a) \) is nonpositive. Thus either \( x_{2m+2} \geq a \) or \( x_{2m-1} \geq a \). But each of these implies that both \( x_{2m+2} \geq -\frac{1}{m+1} \) and \( x_{2m+3} \geq -\frac{1}{m+1} \). It follows that \( -\frac{2}{n} < x_n < 0 \) for all \( n \) and then that \( \lim_{n \to \infty} x_n = \boxed{0} \). | numerical | putnam | Algebra Analysis | The sequence \( \{Q_n(x)\} \) of polynomials is defined by
\[ Q_1(x) = 1 + x, \quad Q_2(x) = 1 + 2x, \]
and, for \( m > 1 \), by
\[ Q_{2m+1}(x) = Q_{2m}(x) + (m+1)xQ_{2m-1}(x), \quad Q_{2m+2}(x) = Q_{2m+1}(x) + (m+1)xQ_{2m}(x). \]
Let \( x_n \) be the largest real solution of \( Q_n(x) = 0 \). Prove that \( \{x_n\} \) is an increasing sequence and that \( \lim_{n \to \infty} x_n = 0 \). | Clearly, \(x_1 = -1, \ x_2 = -\frac{1}{2} \). An easy induction shows that each \( Q_n \) is positive for \(x \geq 0\). Hence \( x_n < 0 \), if \( Q_n \) has zeros.
Assume inductively that \( x_1 < x_2 < \cdots < x_{2m-1} < x_{2m} \). Then \( Q_{2m-1}(x) > 0 \) for \( x > x_{2m-1} \). In particular,
\[ Q_{2m+1}(x_{2m}) = Q_{2m}(x_{2m}) + (m+1)x_{2m}Q_{2m-1}(x_{2m}) < 0. \]
This implies that \( Q_{2m+1}(x) = 0 \) for some \( x > x_{2m} \), i.e., \( x_{2m+1} > x_{2m} \). Similarly, one shows that \( x_{2m+2} > x_{2m+1} \).
Let \( a = -\frac{1}{m+1} \). Using the given recursive definition of the \( Q_n(x) \), one finds that
\[ Q_{2m+2}(a) = Q_{2m+1}(a) - Q_{2m}(a) = -Q_{2m-1}(a). \]
Hence at least one of \( Q_{2m+2}(a) \) and \( Q_{2m-1}(a) \) is nonpositive. Thus either \( x_{2m+2} \geq a \) or \( x_{2m-1} \geq a \). But each of these implies that both \( x_{2m+2} \geq -\frac{1}{m+1} \) and \( x_{2m+3} \geq -\frac{1}{m+1} \). It follows that \( -\frac{2}{n} < x_n < 0 \) for all \( n \) and then that \( \lim_{n \to \infty} x_n = 0 \). | 0 |
1978 | 1978_B5 | Find the largest $A$ for which there exists a polynomial \[ P(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E, \] with real coefficients, which satisfies \[ 0 < P(x) < 1 \quad \text{for } -1 < x < 1. \] | The solution is very easy if one knows that the Chebyshev polynomial \[ C(x) = 8x^4 - 8x^2 + 1 = \cos(4 \arccos x) \] has the largest leading coefficient of all fourth degree polynomials $f(x)$ satisfying \[ -1 < f(x) < 1 \text{ for } -1 < x < 1; \] then one lets \[ P(x) = [C(x) + 1]/2 \] and has $4$ as the largest $A$. Without this information, one can use various substitutions to change the problem into equivalent ones of maximizing $A$ in simpler functions satisfying conditions over intervals. With \[ Q(x) = [P(x) + P(-x)]/2, \] the condition becomes \[ 0 < Q(x) = Ax^4 + Cx^2 + E < 1 \quad \text{over } -1 < x < 1. \] Letting $x^2 = y$, this becomes \[ 0 < R(y) = Ay^2 + Cy + E < 1 \quad \text{over } 0 < y < 1. \] Letting $y = (z + 1)/2$ and $S(z) = R[(z + 1)/2]$, one has \[ 0 < S(z) = (A/4)z^2 + Fz + G < 1 \quad \text{over } -1 < z < 1. \] With \[ T(z) = [S(z) + S(-z)]/2, \] one obtains \[ 0 < (A/4)z^2 + G < 1 \quad \text{over } -1 < z < 1. \] Finally, letting $z^2 = w$, it becomes \[ 0 < (A/4)w + G < 1 \quad \text{over } 0 < w < 1. \] Now it is clear that the maximum $A$ is $4$ and that this maximum is achieved with $G = 0$, i.e., with \[ T(z) = z^2, \quad R(y) = (2y - 1)^2, \quad Q(x) = 4x^4 - 4x^2 + 1. \] Thus the largest A found is \boxed{4}. | numerical | putnam | Algebra Analysis | Find the largest $A$ for which there exists a polynomial \[ P(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E, \] with real coefficients, which satisfies \[ 0 < P(x) < 1 \quad \text{for } -1 < x < 1. \] | The solution is very easy if one knows that the Chebyshev polynomial \[ C(x) = 8x^4 - 8x^2 + 1 = \cos(4 \arccos x) \] has the largest leading coefficient of all fourth degree polynomials $f(x)$ satisfying \[ -1 < f(x) < 1 \text{ for } -1 < x < 1; \] then one lets \[ P(x) = [C(x) + 1]/2 \] and has $\boxed{4}$ as the largest $A$. Without this information, one can use various substitutions to change the problem into equivalent ones of maximizing $A$ in simpler functions satisfying conditions over intervals. With \[ Q(x) = [P(x) + P(-x)]/2, \] the condition becomes \[ 0 < Q(x) = Ax^4 + Cx^2 + E < 1 \quad \text{over } -1 < x < 1. \] Letting $x^2 = y$, this becomes \[ 0 < R(y) = Ay^2 + Cy + E < 1 \quad \text{over } 0 < y < 1. \] Letting $y = (z + 1)/2$ and $S(z) = R[(z + 1)/2]$, one has \[ 0 < S(z) = (A/4)z^2 + Fz + G < 1 \quad \text{over } -1 < z < 1. \] With \[ T(z) = [S(z) + S(-z)]/2, \] one obtains \[ 0 < (A/4)z^2 + G < 1 \quad \text{over } -1 < z < 1. \] Finally, letting $z^2 = w$, it becomes \[ 0 < (A/4)w + G < 1 \quad \text{over } 0 < w < 1. \] Now it is clear that the maximum $A$ is $4$ and that this maximum is achieved with $G = 0$, i.e., with \[ T(z) = z^2, \quad R(y) = (2y - 1)^2, \quad Q(x) = 4x^4 - 4x^2 + 1. \] | 0 |
1979 | 1979_A1 | Find positive integers $n$ and $a_1, a_2, \ldots, a_n$ such that \[ a_1 + a_2 + \cdots + a_n = 1979 \] and the product $a_1 a_2 \cdots a_n$ is as large as possible. Give $n$ as the final answer. | We see that $n = 660$ and that all but one of the $a_i$ equal 3 and the exceptional $a_i$ is a 2 as follows. No $a_i$ can be greater than 4 since one could increase the product by replacing 5 by $2 \cdot 3$, 6 by $3 \cdot 3$, 7 by $3 \cdot 4$, etc. There cannot be both a 2 and a 4 or three 2's among the $a_i$ since $2 \cdot 4 < 3 \cdot 3$ and $2 \cdot 2 \cdot 2 < 3 \cdot 3$. Also, there cannot be two 4's since $4 \cdot 4 < 2 \cdot 3 \cdot 3$. Clearly, no $a_i$ is a 1. Hence the $a_i$ are 3's except possibly for a 4 or for a 2 or for two 2's. Since $1979 = 3 \cdot 659 + 2$, the only exception is a 2 and $n = \boxed{660}$. | numerical | putnam (modified boxing) | Number Theory Combinatorics | Find positive integers $n$ and $a_1, a_2, \ldots, a_n$ such that \[ a_1 + a_2 + \cdots + a_n = 1979 \] and the product $a_1 a_2 \cdots a_n$ is as large as possible. | We see that $n = 660$ and that all but one of the $a_i$ equal 3 and the exceptional $a_i$ is a 2 as follows. No $a_i$ can be greater than 4 since one could increase the product by replacing 5 by $2 \cdot 3$, 6 by $3 \cdot 3$, 7 by $3 \cdot 4$, etc. There cannot be both a 2 and a 4 or three 2's among the $a_i$ since $2 \cdot 4 < 3 \cdot 3$ and $2 \cdot 2 \cdot 2 < 3 \cdot 3$. Also, there cannot be two 4's since $4 \cdot 4 < 2 \cdot 3 \cdot 3$. Clearly, no $a_i$ is a 1. Hence the $a_i$ are 3's except possibly for a 4 or for a 2 or for two 2's. Since $\boxed{1979 = 3 \cdot 659 + 2}$, the only exception is a 2 and $n = 660$. | 0 |
1979 | 1979_B2 | Let $0 < a < b$. Evaluate \[ \lim_{t \to 0} \left( \int_0^1 [bx + a(1-x)]^t \, dx \right)^{1/t}. \] [The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.] | Let $u = bx + a(1-x)$; then the definite integral becomes \[ I(t) = \frac{1}{b-a} \int_a^b u^t \, du = \frac{b^{t+1} - a^{t+1}}{(1+t)(b-a)}. \] Using standard calculus methods for evaluating limits of indeterminate expressions, one finds that \[ \left[ I(t) \right]^{1/t} \to \boxed{e^{-1} (b^b / a^a)^{1/(b-a)}} \text{ as } t \to 0. \] | algebraic | putnam | Calculus Analysis | Let $0 < a < b$. Evaluate \[ \lim_{t \to 0} \left( \int_0^1 [bx + a(1-x)]^t \, dx \right)^{1/t}. \] [The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.] | Let $u = bx + a(1-x)$; then the definite integral becomes \[ I(t) = \frac{1}{b-a} \int_a^b u^t \, du = \frac{b^{t+1} - a^{t+1}}{(1+t)(b-a)}. \] Using standard calculus methods for evaluating limits of indeterminate expressions, one finds that \[ \left[ I(t) \right]^{1/t} \to \boxed{e^{-1} (b^b / a^a)^{1/(b-a)}} \text{ as } t \to 0. \] | 0 |
1979 | 1979_B3 | Let $F$ be a finite field having an odd number $m$ of elements. Let $p(x)$ be an irreducible (i.e., nonfactorable) polynomial over $F$ of the form
\[ x^2 + bx + c, \quad b, c \in F. \]
For how many elements $k$ in $F$ is $p(x) + k$ irreducible over $F$? | Let $r = (m - 1)/2$. We show that $q(x) = p(x) + k$ is irreducible over $F$ for $r$ elements $k$ of $F$. Since $m$ is odd, the characteristic of $F$ is not 2, $1 + 1 = 2 \neq 0$, $2^{-1}b$ is an element $h$ of $F$, the $2r + 1$ elements of $F$ can be expressed in the form $0, f_1, -f_1, \dots, f_r, -f_r$, and $\{0, f_1^2, \dots, f_r^2\}$ is the set of the $r + 1$ distinct squares in $F$. Now
\[ q(x) = (x + h)^2 - (h^2 - c - k) \]
is irreducible over $F$ if and only if it has no zero in $F$, i.e., if and only if $h^2 - c - k$ is not one of the $r + 1$ squares $f^2$ in $F$. Hence $k$ must be one of the $r$ elements left when the $r + 1$ elements of the form $h^2 - c - f^2$ are removed from the $2r + 1$ elements of $F$. Thus the number of elements $k$ in $F$ for which $q(x)$ is irreducible is $r = \boxed{(m - 1)/2}$. | algebraic | putnam | Algebra Number Theory | Let $F$ be a finite field having an odd number $m$ of elements. Let $p(x)$ be an irreducible (i.e., nonfactorable) polynomial over $F$ of the form
\[ x^2 + bx + c, \quad b, c \in F. \]
For how many elements $k$ in $F$ is $p(x) + k$ irreducible over $F$? | Let $r = \boxed{(m - 1)/2}$. We show that $q(x) = p(x) + k$ is irreducible over $F$ for $r$ elements $k$ of $F$. Since $m$ is odd, the characteristic of $F$ is not 2, $1 + 1 = 2 \neq 0$, $2^{-1}b$ is an element $h$ of $F$, the $2r + 1$ elements of $F$ can be expressed in the form $0, f_1, -f_1, \dots, f_r, -f_r$, and $\{0, f_1^2, \dots, f_r^2\}$ is the set of the $r + 1$ distinct squares in $F$. Now
\[ q(x) = (x + h)^2 - (h^2 - c - k) \]
is irreducible over $F$ if and only if it has no zero in $F$, i.e., if and only if $h^2 - c - k$ is not one of the $r + 1$ squares $f^2$ in $F$. Hence $k$ must be one of the $r$ elements left when the $r + 1$ elements of the form $h^2 - c - f^2$ are removed from the $2r + 1$ elements of $F$. | 0 |
1979 | 1979_B4 | (a) Find a solution that is not identically zero, of the homogeneous linear differential equation
\[ (3x^2 + x - 1)y'' - (9x^2 + 9x - 2)y' + (18x + 3)y = 0. \]
Intelligent guessing of the form of a solution may be helpful.
(b) Let $y = f(x)$ be the solution of the nonhomogeneous differential equation
\[ (3x^2 + x - 1)y'' - (9x^2 + 9x - 2)y' + (18x + 3)y = 6(6x + 1) \]
that has $f(0) = 1$ and $(f(-2) - a)(f(2) - b) = c$. Find integers $a, b, c$ such that $(f(-2) - a)(f(2) - b) = c.$ Give the final answer as a sum of the solution from (a) and the values of $a$, $b$ and $c$ from (b). | (a) Trial of $e^{3x}$ shows that $y = e^{3x}$ satisfies the homogeneous equation. Trial of a polynomial $x^d + \dots$ shows that $d$ must be $2$ and then trial of $x^2 + px + q$ shows that $y = x^2 + x$ is a solution. Any linear combination $he^{3x} + k(x^2 + x)$, with at least one of the constants $h$ and $k$ not zero, is an answer.
(b) It is easy to see that $y = 2$ satisfies the nonhomogeneous equation and hence $f(x)$ is of the form $2 + he^{3x} + k(x^2 + x)$. Now $f(0) = 1$ gives us $2 + h = 1$ or $h = -1$. Then $[(f(-1) - 2)][f(1) - 6] = 1$ leads to
\[-e^{-3}(2 + 2k - e^3 - 6) = 1, \quad (2k - 4)e^{-3} = 0, \quad k = 2. \]
Hence $f(x) = 2 - e^{3x} + 2(x^2 + x)$, $f(-2) = 6 - e^{-6}$, $f(2) = 14 - e^6$. Therefore we let $a = 6, b = 14, c = 1.$
We note that if one stops guessing after obtaining an answer $g(x)$ to (a), the standard substitutions $y = g(x)z$ followed by $z' = w$ will reduce the nonhomogeneous equation to a linear equation which can be solved by a well-known method. Having both (a) and (b) we see the final answer is \boxed{e^{3x} + 21}. | algebraic | putnam (modified boxing) | Differential Equations | (a) Find a solution that is not identically zero, of the homogeneous linear differential equation
\[ (3x^2 + x - 1)y'' - (9x^2 + 9x - 2)y' + (18x + 3)y = 0. \]
Intelligent guessing of the form of a solution may be helpful.
(b) Let $y = f(x)$ be the solution of the nonhomogeneous differential equation
\[ (3x^2 + x - 1)y'' - (9x^2 + 9x - 2)y' + (18x + 3)y = 6(6x + 1) \]
that has $f(0) = 1$ and $(f(-2) - a)(f(2) - b) = c$. Find integers $a, b, c$ such that $(f(-2) - a)(f(2) - b) = c.$ | (a) Trial of $e^{3x}$ shows that $y = e^{3x}$ satisfies the homogeneous equation. Trial of a polynomial $x^d + \dots$ shows that $d$ must be $2$ and then trial of $x^2 + px + q$ shows that $y = x^2 + x$ is a solution. Any linear combination $he^{3x} + k(x^2 + x)$, with at least one of the constants $h$ and $k$ not zero, is an answer.
(b) It is easy to see that $y = 2$ satisfies the nonhomogeneous equation and hence $f(x)$ is of the form $2 + he^{3x} + k(x^2 + x)$. Now $f(0) = 1$ gives us $2 + h = 1$ or $h = -1$. Then $[(f(-1) - 2)][f(1) - 6] = 1$ leads to
\[-e^{-3}(2 + 2k - e^3 - 6) = 1, \quad (2k - 4)e^{-3} = 0, \quad k = 2. \]
Hence $f(x) = 2 - e^{3x} + 2(x^2 + x)$, $f(-2) = 6 - e^{-6}$, $f(2) = 14 - e^6$. Therefore we let $a = 6, b = 14, c = 1.$
We note that if one stops guessing after obtaining an answer $g(x)$ to (a), the standard substitutions $y = g(x)z$ followed by $z' = w$ will reduce the nonhomogeneous equation to a linear equation which can be solved by a well-known method. | 0 |
1979 | 1979_B5 | In the plane, let $C$ be a closed convex set that contains $(0,0)$ but no other point with integer coordinates. Suppose that $A(C)$, the area of $C$, is equally distributed among the four quadrants. Find the maximum value of $A(C)$. | A support line for $C$ is a straight line touching $C$ such that one side of the line has no points of $C$. There is a support line containing $(0,1)$; let its slope be $m$. If $m > 1/2$, the part of the area of $C$ in the fourth quadrant is no more than 1 and we are done. Similarly, if $m < -1/2$. So we assume that $-1/2 < m < 1/2$ and assume the analogous facts for support lines containing $(1,0)$, $(0,-1)$, and $(-1,0)$. At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle $\alpha$ to be at a vertex $(h,k)$ in the first quadrant. Then $\alpha \geq \pi/2$ implies that $h+k \leq 2$, and this in turn implies that the area of $C$ in the first quadrant does not exceed 1. Hence $A(C) \leq \boxed{4}$. | numerical | putnam (modified boxing) | Geometry | In the plane, let $C$ be a closed convex set that contains $(0,0)$ but no other point with integer coordinates. Suppose that $A(C)$, the area of $C$, is equally distributed among the four quadrants. Prove that $A(C) \leq 4$. | A support line for $C$ is a straight line touching $C$ such that one side of the line has no points of $C$. There is a support line containing $(0,1)$; let its slope be $m$. If $m > 1/2$, the part of the area of $C$ in the fourth quadrant is no more than 1 and we are done. Similarly, if $m < -1/2$. So we assume that $-1/2 < m < 1/2$ and assume the analogous facts for support lines containing $(1,0)$, $(0,-1)$, and $(-1,0)$. At least one of the angles of the quadrilateral formed by these four support lines is not acute; we may take this angle $\alpha$ to be at a vertex $(h,k)$ in the first quadrant. Then $\alpha \geq \pi/2$ implies that $h+k \leq 2$, and this in turn implies that the area of $C$ in the first quadrant does not exceed 1. Hence $A(C) \leq 4$. | 0 |
1980 | 1980_A1 | Let $b$ and $c$ be fixed real numbers and let the ten points $(j, y_j), j = 1, 2, \dots, 10,$ lie on the parabola $y = x^2 + bx + c$. For $j = 1, 2, \dots, 9$, let $I_j$ be the point of intersection of the tangents to the given parabola at $(j, y_j)$ and $(j+1, y_{j+1})$. Determine the polynomial function $y = g(x)$ of least degree whose graph passes through all nine points $I_j$. | The equation of the tangent to the given parabola at $P_j = (j, y_j)$ is easily seen to be $y = L_j$, where $L_j = (2j + b)x - j^2 + c$. Solving $y = L_j$ and $y = L_{j+1}$ simultaneously, one finds that $x = (2j + 1)/2$ and so $j = (2x - 1)/2$ at $I_j$. Substituting this expression for $j$ into $L_j$ gives the $g(x)$ above. Thus we show that $g(x) = \boxed{x^2 + bx + c - (1/4)}$. | algebraic | putnam | Algebra Geometry | Let $b$ and $c$ be fixed real numbers and let the ten points $(j, y_j), j = 1, 2, \dots, 10,$ lie on the parabola $y = x^2 + bx + c$. For $j = 1, 2, \dots, 9$, let $I_j$ be the point of intersection of the tangents to the given parabola at $(j, y_j)$ and $(j+1, y_{j+1})$. Determine the polynomial function $y = g(x)$ of least degree whose graph passes through all nine points $I_j$. | We show that $g(x) = \boxed{x^2 + bx + c - (1/4)}$. The equation of the tangent to the given parabola at $P_j = (j, y_j)$ is easily seen to be $y = L_j$, where $L_j = (2j + b)x - j^2 + c$. Solving $y = L_j$ and $y = L_{j+1}$ simultaneously, one finds that $x = (2j + 1)/2$ and so $j = (2x - 1)/2$ at $I_j$. Substituting this expression for $j$ into $L_j$ gives the $g(x)$ above. | 0 |
1980 | 1980_A2 | Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a, b, c, d)$ of positive integers such that \[3^r \cdot 7^s = \text{lcm}[a, b, c] = \text{lcm}[a, b, d] = \text{lcm}[a, c, d] = \text{lcm}[b, c, d].\] The answer should be a function of $r$ and $s$. | Each of $a, b, c, d$ must be of the form $3^m 7^n$ with $m$ in $\{0, 1, \ldots, r\}$ and $n$ in $\{0, 1, \ldots, s\}$. Also $m$ must be $r$ for at least two of the four numbers, and $n$ must be $s$ for at least two of the four numbers. There is one way to have $m = r$ for all four numbers, $4r$ ways to have one $m$ in $\{0, 1, \ldots, r-1\}$ and the other three equal to $r$, and $\binom{4}{2}r^2 = 6r^2$ ways to have two of the $m$'s in $\{0, 1, \ldots, r-1\}$ and the other two equal to $r$. Thus there are $1 + 4r + 6r^2$ choices of allowable $m$'s and, similarly, $1 + 4s + 6s^2$ choices of allowable $n$'s. Thus we show that the number is $\boxed{(1 + 4r + 6r^2)(1 + 4s + 6s^2)}$. | algebraic | putnam | Algebra Number Theory | Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a, b, c, d)$ of positive integers such that \[3^r \cdot 7^s = \text{lcm}[a, b, c] = \text{lcm}[a, b, d] = \text{lcm}[a, c, d] = \text{lcm}[b, c, d].\] The answer should be a function of $r$ and $s$. | We show that the number is $\boxed{(1 + 4r + 6r^2)(1 + 4s + 6s^2)}$. Each of $a, b, c, d$ must be of the form $3^m 7^n$ with $m$ in $\{0, 1, \ldots, r\}$ and $n$ in $\{0, 1, \ldots, s\}$. Also $m$ must be $r$ for at least two of the four numbers, and $n$ must be $s$ for at least two of the four numbers. There is one way to have $m = r$ for all four numbers, $4r$ ways to have one $m$ in $\{0, 1, \ldots, r-1\}$ and the other three equal to $r$, and $\binom{4}{2}r^2 = 6r^2$ ways to have two of the $m$'s in $\{0, 1, \ldots, r-1\}$ and the other two equal to $r$. Thus there are $1 + 4r + 6r^2$ choices of allowable $m$'s and, similarly, $1 + 4s + 6s^2$ choices of allowable $n$'s. | 0 |
1980 | 1980_A3 | Evaluate \[ \int_0^{\pi/2} \frac{dx}{1 + (\tan x)^\sqrt{2}}. \] | Let $I$ be the given definite integral and $\sqrt{2} = r$. We show that $I = \pi/4$. Using $x = (\pi/2) - u$, one has \[ I = \int_0^{\pi/2} \frac{\tan' u \; du}{\tan' u + 1}. \] Hence \[ 2I = \int_0^{\pi/2} \frac{1 + \tan^r x}{1 + \tan^r x} dx = \pi/2 \; \text{and} \; I = \boxed{\pi/4}. \] | numerical | putnam | Calculus | Evaluate \[ \int_0^{\pi/2} \frac{dx}{1 + (\tan x)^\sqrt{2}}. \] | Let $I$ be the given definite integral and $\sqrt{2} = r$. We show that $I = \pi/4$. Using $x = (\pi/2) - u$, one has \[ I = \int_0^{\pi/2} \frac{\tan' u \; du}{\tan' u + 1}. \] Hence \[ 2I = \int_0^{\pi/2} \frac{1 + \tan^r x}{1 + \tan^r x} dx = \pi/2 \; \text{and} \; I = \boxed{\pi/4}. \] | 0 |
1980 | 1980_A6 | Let $C$ be the class of all real valued continuously differentiable functions $f$ on the interval $0 < x < 1$ with $f(0) = 0$ and $f(1) = 1$. Determine the largest real number $u$ such that \[ u \leq \int_0^1 |f'(x) - f(x)| dx \] for all $f$ in $C$. | We show that $u = 1/e$. Since $f' - f = (fe^{-x})'e^x$ and $e^x \geq 1$ for $x \geq 0$, \[ \int_0^1 |f' - f| dx = \int_0^1 |(fe^{-x})'e^x| dx \geq \int_0^1 (fe^{-x})' dx = [fe^{-x}]_0^1 = 1/e. \] To see that $1/e$ is the largest lower bound, we use functions $f_a(x)$ defined by \[ f_a(x) = (e^{a-1}/a)x \text{ for } 0 \leq x \leq a, \quad f_a(x) = e^{x-1} \text{ for } a \leq x \leq 1. \] Let $m = e^{a-1}/a$. Then \[ \int_0^1 |f'_a(x) - f_a(x)| dx = \int_0^a |m - mx| dx = m \left( a - \frac{a^2}{2} \right) = e^{a-1} \left( 1 - \frac{a}{2} \right). \] As $a \to 0$, this expression approaches $1/e$. The function $f_a(x)$ does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than $\boxed{1/e}$ can be an upper bound. | algebraic | putnam | Analysis Calculus | Let $C$ be the class of all real valued continuously differentiable functions $f$ on the interval $0 < x < 1$ with $f(0) = 0$ and $f(1) = 1$. Determine the largest real number $u$ such that \[ u \leq \int_0^1 |f'(x) - f(x)| dx \] for all $f$ in $C$. | We show that $u = 1/e$. Since $f' - f = (fe^{-x})'e^x$ and $e^x \geq 1$ for $x \geq 0$, \[ \int_0^1 |f' - f| dx = \int_0^1 |(fe^{-x})'e^x| dx \geq \int_0^1 (fe^{-x})' dx = [fe^{-x}]_0^1 = 1/e. \] To see that $1/e$ is the largest lower bound, we use functions $f_a(x)$ defined by \[ f_a(x) = (e^{a-1}/a)x \text{ for } 0 \leq x \leq a, \quad f_a(x) = e^{x-1} \text{ for } a \leq x \leq 1. \] Let $m = e^{a-1}/a$. Then \[ \int_0^1 |f'_a(x) - f_a(x)| dx = \int_0^a |m - mx| dx = m \left( a - \frac{a^2}{2} \right) = e^{a-1} \left( 1 - \frac{a}{2} \right). \] As $a \to 0$, this expression approaches $1/e$. The function $f_a(x)$ does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than $\boxed{1/e}$ can be an upper bound. | 0 |
1980 | 1980_B1 | Find the lower bound for which real numbers $c$ is \[(e^x + e^{-x})/2 \leq e^{cx^2}\] for all real $x$? | The inequality holds if and only if $c \geq 1/2$. For $c \geq 1/2$, \[\frac{e^x + e^{-x}}{2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} < \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n n!} = e^{x^2 / 2} \leq e^{cx^2}\] for all $x$ since $(2n)! > 2^n n!$ for $n = 0, 1, \ldots$. Conversely, if the inequality holds for all $x$, then \[0 \leq \lim_{x \to 0} \frac{e^{cx^2} - \frac{1}{2}(e^x + e^{-x})}{x^2} = \lim_{x \to 0} \frac{(1 + cx^2 + \ldots) - (1 + \frac{1}{2}x^2 + \ldots)}{x^2} = c - \frac{1}{2}\] and so $c \geq \boxed{1/2}$. | numerical | putnam (modified boxing) | Algebra Analysis | For which real numbers $c$ is \[(e^x + e^{-x})/2 \leq e^{cx^2}\] for all real $x$? | The inequality holds if and only if $c \geq 1/2$. For $c \geq 1/2$, \[\frac{e^x + e^{-x}}{2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} < \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n n!} = e^{x^2 / 2} \leq e^{cx^2}\] for all $x$ since $(2n)! > 2^n n!$ for $n = 0, 1, \ldots$. Conversely, if the inequality holds for all $x$, then \[0 \leq \lim_{x \to 0} \frac{e^{cx^2} - \frac{1}{2}(e^x + e^{-x})}{x^2} = \lim_{x \to 0} \frac{(1 + cx^2 + \ldots) - (1 + \frac{1}{2}x^2 + \ldots)}{x^2} = c - \frac{1}{2}\] and so $\boxed{c \geq 1/2}$. | 0 |
1980 | 1980_B2 | Let $S$ be the solid in three-dimensional space consisting of all points $(x, y, z)$ satisfying the following system of six simultaneous conditions:
\[\begin{aligned}
x &\geq 0, \quad y \geq 0, \quad z \geq 0, \\
x + y + z &\leq 11, \\
2x + 4y + 3z &\leq 36, \\
2x + 3z &\leq 24.
\end{aligned}\]
\begin{enumerate}
\item[(a)] Determine the number $v$ of vertices of $S$.
\item[(b)] Determine the number $e$ of edges of $S$.
\end{enumerate} Return the sum of both numbers. | (a) $v = 7$. The seven vertices are $V_0 = (0, 0, 0)$, $V_1 = (11, 0, 0)$, $V_2 = (0, 9, 0)$, $V_3 = (0, 0, 8)$, $V_4 = (0, 3, 8)$, $V_5 = (9, 0, 2)$, and $V_6 = (4, 7, 0)$.
(b) $e = 11$. The eleven edges are $V_0V_1$, $V_0V_2$, $V_0V_3$, $V_1V_5$, $V_1V_6$, $V_2V_4$, $V_2V_6$, $V_3V_4$, $V_3V_5$, $V_4V_5$, and $V_4V_6$. Thus the final answer is $\boxed{18}$. | numerical | putnam (modified boxing) | Geometry | Let $S$ be the solid in three-dimensional space consisting of all points $(x, y, z)$ satisfying the following system of six simultaneous conditions:
\[\begin{aligned}
x &\geq 0, \quad y \geq 0, \quad z \geq 0, \\
x + y + z &\leq 11, \\
2x + 4y + 3z &\leq 36, \\
2x + 3z &\leq 24.
\end{aligned}\]
\begin{enumerate}
\item[(a)] Determine the number $v$ of vertices of $S$.
\item[(b)] Determine the number $e$ of edges of $S$.
\item[(c)] Sketch in the $bc$-plane the set of points $(b, c)$ such that $(2, 5, 4)$ is one of the points $(x, y, z)$ at which the linear function $bx + cy + z$ assumes its maximum value on $S$.
\end{enumerate} | (a) $v = 7$. The seven vertices are $V_0 = (0, 0, 0)$, $V_1 = (11, 0, 0)$, $V_2 = (0, 9, 0)$, $V_3 = (0, 0, 8)$, $V_4 = (0, 3, 8)$, $V_5 = (9, 0, 2)$, and $V_6 = (4, 7, 0)$.
(b) $e = 11$. The eleven edges are $V_0V_1$, $V_0V_2$, $V_0V_3$, $V_1V_5$, $V_1V_6$, $V_2V_4$, $V_2V_6$, $V_3V_4$, $V_3V_5$, $V_4V_5$, and $V_4V_6$.
(c) The desired $(b, c)$ are those with $b + c = 2$ and $\frac{2}{3} \leq b \leq 1$. Let $L(x, y, z) = bx + cy + z$. Since $L$ is linear and $(2, 5, 4)$ is on edge $V_4V_6$, the maximum of $L$ on $S$ must be assumed at $V_4$ and $V_6$, and the conditions on $b$ and $c$ are obtained from $L(0, 3, 8) = L(4, 7, 0) \geq L(x, y, z)$, with $(x, y, z)$ ranging over the other five vertices. | 0 |
1980 | 1980_B3 | For what lower bound of real numbers $a$ does the sequence defined by the initial condition $u_0 = a$ and the recursion $u_{n+1} = 2u_n - n^2$ have $u_n > 0$ for all $n \geq 0$?
(Express the answer in the simplest form.) | We show that $u_n > 0$ for all $n \geq 0$ if and only if $a \geq 3$. Let $\Delta u_n = u_{n+1} - u_n$. Then the recursion (i.e., difference equation) takes the form $(1 - \Delta) u_n = n^2$. Since $n^2$ is a polynomial, a particular solution is
\[u_n = (1 - \Delta)^{-1}n^2 = (1 + \Delta + \Delta^2 + \cdots)n^2 = n^2 + (2n + 1) + 2 = n^2 + 2n + 3.\]
(This is easily verified by substitution.) The complete solution is $u_n = n^2 + 2n + 3 + k\cdot 2^n$, since $v_n = k\cdot 2^n$ is the solution of the associated homogeneous difference equation $v_{n+1} - 2v_n = 0$. The desired solution with $u_0 = a$ is $u_n = n^2 + 2n + 3 + (a - 3)2^n$. Since $\lim_{n \to \infty}[2^n/(n^2 + 2n + 3)] = +\infty$, $u_n$ will be negative for large enough $n$ if $a - 3 < 0$. Conversely, if $a - 3 \geq 0$, it is clear that each $u_n > 0$ making the final answer $\boxed{3}$. | numerical | putnam (modified boxing) | Algebra | For which real numbers $a$ does the sequence defined by the initial condition $u_0 = a$ and the recursion $u_{n+1} = 2u_n - n^2$ have $u_n > 0$ for all $n \geq 0$?
(Express the answer in the simplest form.) | We show that $u_n > 0$ for all $n \geq 0$ if and only if $\boxed{a \geq 3}$. Let $\Delta u_n = u_{n+1} - u_n$. Then the recursion (i.e., difference equation) takes the form $(1 - \Delta) u_n = n^2$. Since $n^2$ is a polynomial, a particular solution is
\[u_n = (1 - \Delta)^{-1}n^2 = (1 + \Delta + \Delta^2 + \cdots)n^2 = n^2 + (2n + 1) + 2 = n^2 + 2n + 3.\]
(This is easily verified by substitution.) The complete solution is $u_n = n^2 + 2n + 3 + k\cdot 2^n$, since $v_n = k\cdot 2^n$ is the solution of the associated homogeneous difference equation $v_{n+1} - 2v_n = 0$. The desired solution with $u_0 = a$ is $u_n = n^2 + 2n + 3 + (a - 3)2^n$. Since $\lim_{n \to \infty}[2^n/(n^2 + 2n + 3)] = +\infty$, $u_n$ will be negative for large enough $n$ if $a - 3 < 0$. Conversely, if $a - 3 \geq 0$, it is clear that each $u_n > 0$. | 0 |
1981 | 1981_A1 | Let $E(n)$ denote the largest integer $k$ such that $5^k$ is an integral divisor of the product $1^1 2^2 3^3 \dots n^n$. Calculate \[ \lim_{n \to \infty} \frac{E(n)}{n^2}. \] | We show that the limit is $1/8$. Let $T(m) = 1 + 2 + \dots + m = m(m+1)/2$, $\lfloor x \rfloor$ denote the greatest integer in $x$, $h = \lfloor \log_5 n \rfloor$, and $e_i$ be the fractional part $(n/5^i) - \lfloor n/5^i \rfloor$ for $1 \leq i \leq h$. Then \[ E(n) = 5T(\lfloor n/5 \rfloor) + 5^2T(\lfloor n/5^2 \rfloor) + \dots + 5^hT(\lfloor n/5^h \rfloor) \] \[ 2E(n) = 5(\lfloor n/5 \rfloor)^2 + \lfloor n/5 \rfloor + 5^2(\lfloor n/5^2 \rfloor)^2 + \lfloor n/5^2 \rfloor + \dots + 5^h(\lfloor n/5^h \rfloor)^2 + \lfloor n/5^h \rfloor \] \[ = 5 \left( \frac{n^2}{5^2} - 2e_1 \frac{n}{5} + e_1^2 \right) + \dots + 5^h \left( \frac{n^2}{5^{2h}} - 2e_h \frac{n}{5^h} + e_h^2 \right). \] \[ \frac{E(n)}{n^2} = \frac{1}{2} \left( \frac{1}{5} + \frac{1}{5^2} + \dots + \frac{1}{5^2} \right) + \frac{h}{2n} - \frac{e_1 + e_2 + \dots + e_h}{n} \] \[ + \frac{5(e_1^2 - e_1) + \dots + 5^h(e_h^2 - e_h)}{2n^2}. \] Since $5^h \leq n < 5^{h+1}$ and $0 \leq e_i < 1$, one sees that $h/n \to 0$ and $E(n)/n^2 \to \boxed{1/8}$ as $n \to \infty$. | numerical | putnam | Number Theory | Let $E(n)$ denote the largest integer $k$ such that $5^k$ is an integral divisor of the product $1^1 2^2 3^3 \dots n^n$. Calculate \[ \lim_{n \to \infty} \frac{E(n)}{n^2}. \] | We show that the limit is $1/8$. Let $T(m) = 1 + 2 + \dots + m = m(m+1)/2$, $\lfloor x \rfloor$ denote the greatest integer in $x$, $h = \lfloor \log_5 n \rfloor$, and $e_i$ be the fractional part $(n/5^i) - \lfloor n/5^i \rfloor$ for $1 \leq i \leq h$. Then \[ E(n) = 5T(\lfloor n/5 \rfloor) + 5^2T(\lfloor n/5^2 \rfloor) + \dots + 5^hT(\lfloor n/5^h \rfloor) \] \[ 2E(n) = 5(\lfloor n/5 \rfloor)^2 + \lfloor n/5 \rfloor + 5^2(\lfloor n/5^2 \rfloor)^2 + \lfloor n/5^2 \rfloor + \dots + 5^h(\lfloor n/5^h \rfloor)^2 + \lfloor n/5^h \rfloor \] \[ = 5 \left( \frac{n^2}{5^2} - 2e_1 \frac{n}{5} + e_1^2 \right) + \dots + 5^h \left( \frac{n^2}{5^{2h}} - 2e_h \frac{n}{5^h} + e_h^2 \right). \] \[ \frac{E(n)}{n^2} = \frac{1}{2} \left( \frac{1}{5} + \frac{1}{5^2} + \dots + \frac{1}{5^2} \right) + \frac{h}{2n} - \frac{e_1 + e_2 + \dots + e_h}{n} \] \[ + \frac{5(e_1^2 - e_1) + \dots + 5^h(e_h^2 - e_h)}{2n^2}. \] Since $5^h \leq n < 5^{h+1}$ and $0 \leq e_i < 1$, one sees that $h/n \to 0$ and $E(n)/n^2 \to \boxed{1/8}$ as $n \to \infty$. | 0 |
1981 | 1981_A2 | Two distinct squares of the 8 by 8 chessboard $C$ are said to be adjacent if they have a vertex or side in common. Also, $g$ is called a $C$-gap if for every numbering of the squares of $C$ with all the integers $1, 2, \dots, 64$ there exist two adjacent squares whose numbers differ by at least $g$. Determine the largest $C$-gap $g$. | For any numbering, one can go from the square numbered $1$ to the square numbered $64$ in $7$ or fewer steps, in each step going to an adjacent square; thus $(64 - 1)/7 = 9$ is a $C$-gap. It is the largest $C$-gap since with coordinates $(a, b)$, $1 \leq a \leq 8$ and $1 \leq b \leq 8$, for the squares we can number $(a, b)$ with $8(a - 1) + b$ and thus find that no number greater than $\boxed{9}$ is a $C$-gap. | numerical | putnam | Combinatorics | Two distinct squares of the 8 by 8 chessboard $C$ are said to be adjacent if they have a vertex or side in common. Also, $g$ is called a $C$-gap if for every numbering of the squares of $C$ with all the integers $1, 2, \dots, 64$ there exist two adjacent squares whose numbers differ by at least $g$. Determine the largest $C$-gap $g$. | For any numbering, one can go from the square numbered $1$ to the square numbered $64$ in $7$ or fewer steps, in each step going to an adjacent square; thus $(64 - 1)/7 = 9$ is a $C$-gap. It is the largest $C$-gap since with coordinates $(a, b)$, $1 \leq a \leq 8$ and $1 \leq b \leq 8$, for the squares we can number $(a, b)$ with $8(a - 1) + b$ and thus find that no number greater than $\boxed{9}$ is a $C$-gap. | 0 |
1981 | 1981_A3 | Find \[ \lim_{t \to \infty} e^{-t} \left[ \int_0^t \int_0^t \frac{e^x - e^y}{x - y} \, dx \, dy \right] \] or show that the limit does not exist by returning the value it approaches as t \to \infty. | Let $G(t)$ be the double integral. Then \[ \lim_{t \to \infty} [G(t)/e^t] = \lim_{t \to \infty} [G'(t)/e^t] \] by L'Hôpital's Rule. One finds that \[ G'(t) = \int_0^t \frac{e^x - e^t}{x - t} \, dx + \int_0^t \frac{e^y - e^t}{y - t} \, dy = 2 \int_0^t \frac{e^x - e^t}{x - t} \, dx. \] Then using $e^x = e^t [1 + (x - t) + (x - t)^2 / 2! + \cdots]$, one sees that $e^{-t} G'(t) \to \boxed{\infty}$ as $t \to \infty$ since for sufficiently large $t$, \[ \frac{G'(t)}{2e^t} = \int_0^t \frac{e^{x-t} - 1}{x - t} \, dx = \int_0^t \frac{1 - e^{-y}}{y} \, dy > \int_1^t \frac{1 - e^{-y}}{y} \, dy > (1 - e^{-1}) \log t. \] | numerical | putnam | Calculus Analysis | Find \[ \lim_{t \to \infty} e^{-t} \left[ \int_0^t \int_0^t \frac{e^x - e^y}{x - y} \, dx \, dy \right] \] or show that the limit does not exist. | Let $G(t)$ be the double integral. Then \[ \lim_{t \to \infty} [G(t)/e^t] = \lim_{t \to \infty} [G'(t)/e^t] \] by L'Hôpital's Rule. One finds that \[ G'(t) = \int_0^t \frac{e^x - e^t}{x - t} \, dx + \int_0^t \frac{e^y - e^t}{y - t} \, dy = 2 \int_0^t \frac{e^x - e^t}{x - t} \, dx. \] Then using $e^x = e^t [1 + (x - t) + (x - t)^2 / 2! + \cdots]$, one sees that $e^{-t} G'(t) \to \infty$ as $t \to \infty$ since for sufficiently large $t$, \[ \frac{G'(t)}{2e^t} = \int_0^t \frac{e^{x-t} - 1}{x - t} \, dx = \int_0^t \frac{1 - e^{-y}}{y} \, dy > \int_1^t \frac{1 - e^{-y}}{y} \, dy > (1 - e^{-1}) \log t. \] | 0 |
1981 | 1981_A4 | A point $P$ moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\nLet $N(T)$ be the number of starting directions from a fixed interior point $P_0$ for which $P$ escapes within $T$ units of time. Find the least constant $a$ for which constants $b$ and $c$ exist such that\n\[ N(T) \leq aT^2 + bT + c \] \nfor all $T > 0$ and all initial points $P_0$. | Set up coordinates so that a vertex of the given unit square is $(0, 0)$ and two sides of the square are on the axes. Using the reflection properties, one can see that $P$ escapes within $T$ units of time if and only if the (infinite) ray from $P_0$, with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within $T$ units of distance from $P_0$. Thus $N(T)$ is at most the number $L(T)$ of lattice points in the circle with center at $P_0$ and radius $T$. Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\[ N(T) \leq L(T) \leq \pi \left[T + (\sqrt{2}/2)\right]^2. \]\nHence there is an upper bound for $N(T)$ of the form $\pi T^2 + bT + c$, with $b$ and $c$ fixed. When just one coordinate of $P_0$ is irrational,\n\[ N(T) = L(T) \geq \pi \left[T - (\sqrt{2}/2)\right]^2. \]\nThis lower bound for $N(T)$ exceeds $aT^2 + bT + c$ for sufficiently large $T$ if $a < \pi$; hence $\boxed{\pi}$ is the desired $a$. | numerical | putnam | Geometry Analysis | A point $P$ moves inside a unit square in a straight line at unit speed. When it meets a corner it escapes. When it meets an edge its line of motion is reflected so that the angle of incidence equals the angle of reflection.\nLet $N(T)$ be the number of starting directions from a fixed interior point $P_0$ for which $P$ escapes within $T$ units of time. Find the least constant $a$ for which constants $b$ and $c$ exist such that\n\[ N(T) \leq aT^2 + bT + c \] \nfor all $T > 0$ and all initial points $P_0$. | Set up coordinates so that a vertex of the given unit square is $(0, 0)$ and two sides of the square are on the axes. Using the reflection properties, one can see that $P$ escapes within $T$ units of time if and only if the (infinite) ray from $P_0$, with the direction of the first segment of the path, goes through a lattice point (point with integer coordinates) within $T$ units of distance from $P_0$. Thus $N(T)$ is at most the number $L(T)$ of lattice points in the circle with center at $P_0$ and radius $T$. Tiling the plane with unit squares having centers at the lattice points and considering areas, one sees that\n\[ N(T) \leq L(T) \leq \pi \left[T + (\sqrt{2}/2)\right]^2. \]\nHence there is an upper bound for $N(T)$ of the form $\pi T^2 + bT + c$, with $b$ and $c$ fixed. When just one coordinate of $P_0$ is irrational,\n\[ N(T) = L(T) \geq \pi \left[T - (\sqrt{2}/2)\right]^2. \]\nThis lower bound for $N(T)$ exceeds $aT^2 + bT + c$ for sufficiently large $T$ if $a < \pi$; hence $\boxed{\pi}$ is the desired $a$. | 0 |
1981 | 1981_A5 | Let $P(x)$ be a polynomial with real coefficients and form the polynomial \[ Q(x) = (x^2+1)P(x)P'(x) + x\left([P(x)]^2 + [P'(x)]^2\right). \] Given that the equation $P(x) = 0$ has $n$ distinct real roots exceeding 1, find the minimum number of distinct real roots the equation $Q(x) = 0$ has. | We show that $Q(x)$ has at least $2n-1$ real zeros. One finds that $Q(x) = F(x)G(x)$, where \[ F(x) = P'(x) + xP(x) = e^{-x^2/2}\left[e^{x^2/2}P(x)\right]', \] \[ G(x) = xP'(x) + P(x) = [xP(x)]'. \] We can assume that $P(x)$ has exactly $n$ zeros $a_i$ exceeding 1 with $1 < a_1 < a_2 < \cdots < a_n$. It follows from Rolle's Theorem that $F(x)$ has $n-1$ zeros $b_i$ and $G(x)$ has $n$ zeros $c_i$ with \[ 1 < a_1 < b_1 < a_2 < b_2 < \cdots < b_{n-1} < a_n, \] \[ 0 < c_1 < a_1 < c_2 < a_2 < \cdots < c_n < a_n. \] If $b_i \neq c_{i+1}$ for all $i$, the $b$'s and $c$'s are $2n-1$ distinct zeros of $Q(x)$. So we assume that $b_i = c_{i+1} = r$ for some $i$. Then \[ P'(r) + rP(r) = 0 = rP'(r) + P(r), \] and so $(r^2-1)P(r) = 0$. Since $r = b_i > 1$, $P(r) = 0$. Since $a_i < r < a_{i+1}$, this contradicts the fact that the $a$'s are all the zeros exceeding 1 of $P(x)$. Hence $Q(x)$ has at least $\boxed{2n-1}$ distinct real zeros. | algebraic | putnam (modified boxing) | Algebra Analysis | Let $P(x)$ be a polynomial with real coefficients and form the polynomial \[ Q(x) = (x^2+1)P(x)P'(x) + x\left([P(x)]^2 + [P'(x)]^2\right). \] Given that the equation $P(x) = 0$ has $n$ distinct real roots exceeding 1, prove or disprove that the equation $Q(x) = 0$ has at least $2n-1$ distinct real roots. | We show that $Q(x)$ has at least $2n-1$ real zeros. One finds that $Q(x) = F(x)G(x)$, where \[ F(x) = P'(x) + xP(x) = e^{-x^2/2}\left[e^{x^2/2}P(x)\right]', \] \[ G(x) = xP'(x) + P(x) = [xP(x)]'. \] We can assume that $P(x)$ has exactly $n$ zeros $a_i$ exceeding 1 with $1 < a_1 < a_2 < \cdots < a_n$. It follows from Rolle's Theorem that $F(x)$ has $n-1$ zeros $b_i$ and $G(x)$ has $n$ zeros $c_i$ with \[ 1 < a_1 < b_1 < a_2 < b_2 < \cdots < b_{n-1} < a_n, \] \[ 0 < c_1 < a_1 < c_2 < a_2 < \cdots < c_n < a_n. \] If $b_i \neq c_{i+1}$ for all $i$, the $b$'s and $c$'s are $2n-1$ distinct zeros of $Q(x)$. So we assume that $b_i = c_{i+1} = r$ for some $i$. Then \[ P'(r) + rP(r) = 0 = rP'(r) + P(r), \] and so $(r^2-1)P(r) = 0$. Since $r = b_i > 1$, $P(r) = 0$. Since $a_i < r < a_{i+1}$, this contradicts the fact that the $a$'s are all the zeros exceeding 1 of $P(x)$. Hence $Q(x)$ has at least $2n-1$ distinct real zeros. | 0 |
1981 | 1981_A6 | Suppose that each of the vertices of $\triangle ABC$ is a lattice point in the $(x, y)$-plane and that there is exactly one lattice point $P$ in the \textit{interior} of the triangle. The line $AP$ is extended to meet $BC$ at $E$. Determine the largest possible value for the ratio of lengths of segments
\[\frac{|AP|}{|PE|}.\]
[A lattice point is a point whose coordinates $x$ and $y$ are integers.] | Treating each point $X$ of the plane as the vector $\overrightarrow{AX}$ with initial point at $A$ and final point at $X$, let
\[ L = \frac{B + C}{2}, \quad M = \frac{C}{2}, \quad \text{and} \quad N = \frac{B}{2} \]
(be the midpoints of sides $BC$, $AC$, and $AB$). Also let
\[ S = \frac{2L + M}{3} = \frac{B + C + M}{3}, \quad T = \frac{2L + N}{3} = \frac{B + C + N}{3}, \quad Q = 2P - B, \quad \text{and} \quad R = 3P - B - C. \]
Clearly $Q$ and $R$ are lattice points. Also $Q \neq P$ and $R \neq P$ since $Q = P$ implies $P$ is the point $L$ on side $BC$. Hence $Q$ is not inside $\triangle NBL$ since the linear transformation $f$ with $f(X) = 2X - B$ translates a doubled $\triangle NBL$ (and its inside) onto $\triangle ABC$ (and its inside). Similarly, $P$ is not inside $\triangle MCL$. Using the mapping $g(X) = 3X - B - C$ and the fact that $R$ is not inside $\triangle LMN$, one finds that $P$ is not inside $\triangle LST$. Since the distance from $A$ to line $ST$ is 5 times the distance between lines $ST$ and $BC$, it follows that
\[ \frac{|AP|}{|PE|} \leq 5. \]
This upper bound 5 is seen to be the maximum by considering the example with $A = (0, 0)$, $B = (0, 2)$, and $C = (3, 0)$ in which $P = (1, 1) = T$ is the only lattice point inside $\triangle ABC$ and $\frac{|AT|}{|TE|} = \boxed{5}. | numerical | putnam | Geometry | Suppose that each of the vertices of $\triangle ABC$ is a lattice point in the $(x, y)$-plane and that there is exactly one lattice point $P$ in the \textit{interior} of the triangle. The line $AP$ is extended to meet $BC$ at $E$. Determine the largest possible value for the ratio of lengths of segments
\[\frac{|AP|}{|PE|}.\]
[A lattice point is a point whose coordinates $x$ and $y$ are integers.] | Treating each point $X$ of the plane as the vector $\overrightarrow{AX}$ with initial point at $A$ and final point at $X$, let
\[ L = \frac{B + C}{2}, \quad M = \frac{C}{2}, \quad \text{and} \quad N = \frac{B}{2} \]
(be the midpoints of sides $BC$, $AC$, and $AB$). Also let
\[ S = \frac{2L + M}{3} = \frac{B + C + M}{3}, \quad T = \frac{2L + N}{3} = \frac{B + C + N}{3}, \quad Q = 2P - B, \quad \text{and} \quad R = 3P - B - C. \]
Clearly $Q$ and $R$ are lattice points. Also $Q \neq P$ and $R \neq P$ since $Q = P$ implies $P$ is the point $L$ on side $BC$. Hence $Q$ is not inside $\triangle NBL$ since the linear transformation $f$ with $f(X) = 2X - B$ translates a doubled $\triangle NBL$ (and its inside) onto $\triangle ABC$ (and its inside). Similarly, $P$ is not inside $\triangle MCL$. Using the mapping $g(X) = 3X - B - C$ and the fact that $R$ is not inside $\triangle LMN$, one finds that $P$ is not inside $\triangle LST$. Since the distance from $A$ to line $ST$ is 5 times the distance between lines $ST$ and $BC$, it follows that
\[ \frac{|AP|}{|PE|} \leq 5. \]
This upper bound 5 is seen to be the maximum by considering the example with $A = (0, 0)$, $B = (0, 2)$, and $C = (3, 0)$ in which $P = (1, 1) = T$ is the only lattice point inside $\triangle ABC$ and $\frac{|AT|}{|TE|} = \boxed{5}. | 0 |
1981 | 1981_B1 | Find
\[\lim_{n \to \infty}\left[ \frac{1}{n^5} \sum_{h=1}^n \sum_{k=1}^n \left(5h^4 - 18h^2k^2 + 5k^4\right) \right].\] | Let $S_k(n) = 1^k + 2^k + \cdots + n^k$. Using standard methods of calculus texts one finds that \[ S_2(n) = \frac{n^3}{3} + \frac{n^2}{2} + an \] and \[ S_4(n) = \frac{n^5}{5} + \frac{n^4}{2} + bn^3 + cn^2 + dn, \] with $a$, $b$, $c$, $d$ constants. Then the double sum is \[ 10nS_4(n) - 18\left[S_2(n)\right]^2 = \left(2n^6 + 5n^5 + \cdots\right) - \left(2n^6 + 6n^5 + \cdots\right) = -n^5 + \cdots \] and the desired limit is \(\boxed{-1}\). | numerical | putnam | Calculus Algebra | Find
\[\lim_{n \to \infty}\left[ \frac{1}{n^5} \sum_{h=1}^n \sum_{k=1}^n \left(5h^4 - 18h^2k^2 + 5k^4\right) \right].\] | Let $S_k(n) = 1^k + 2^k + \cdots + n^k$. Using standard methods of calculus texts one finds that \[ S_2(n) = \frac{n^3}{3} + \frac{n^2}{2} + an \] and \[ S_4(n) = \frac{n^5}{5} + \frac{n^4}{2} + bn^3 + cn^2 + dn, \] with $a$, $b$, $c$, $d$ constants. Then the double sum is \[ 10nS_4(n) - 18\left[S_2(n)\right]^2 = \left(2n^6 + 5n^5 + \cdots\right) - \left(2n^6 + 6n^5 + \cdots\right) = -n^5 + \cdots \] and the desired limit is \(\boxed{-1}\). | 0 |
1981 | 1981_B2 | Determine the minimum value of \[ (r - 1)^2 + \left(\frac{s}{r} - 1\right)^2 + \left(\frac{t}{s} - 1\right)^2 + \left(\frac{4}{t} - 1\right)^2 \] for all real numbers $r, s, t$ with $1 \leq r \leq s \leq t \leq 4$. | First we let $0 < a < b$ and seek the $x$ that minimizes \[ f(x) = \left(\frac{x}{a} - 1\right)^2 + \left(\frac{b}{x} - 1\right)^2 \] on $a \leq x \leq b$. Let $x/a = z$ and $b/a = c$. Then \[ f(x) = g(z) = (z - 1)^2 + \left(\frac{c}{z} - 1\right)^2. \] Now $g'(z) = 0$ implies \[ z^4 - z^3 + cz - c^2 = (z^2 - c)(z^2 - z + c) = 0; \] the only positive solution is $z = \sqrt{c}$. Since $0 < a < b$, $c > 1$, $\sqrt{c} > 1$, and \[ g(1) = g(c) = (c - 1)^2 + (\sqrt{c} - 1)^2(\sqrt{c} + 1)^2 > 2(\sqrt{c} - 1). \] Hence the minimum of $g(z)$ on $1 \leq z \leq c$ occurs at $z = \sqrt{c}$. It follows that the minimum for $f(x)$ on $a \leq x \leq b$ occurs at $x = a\sqrt{b/a} = \sqrt{ab}$. Then the minimum for the given function of $r, s, t$ occurs with $r = \sqrt{5}, t = \sqrt{4s}, r = \sqrt{2}$. These imply that $r = \sqrt{2}, s = 2, t = 2\sqrt{2}$. Thus the desired minimum value is \[ 4(\sqrt{2} - 1)^2 = \boxed{12 - 8\sqrt{2}}. \] | numerical | putnam | Calculus Analysis | Determine the minimum value of \[ (r - 1)^2 + \left(\frac{s}{r} - 1\right)^2 + \left(\frac{t}{s} - 1\right)^2 + \left(\frac{4}{t} - 1\right)^2 \] for all real numbers $r, s, t$ with $1 \leq r \leq s \leq t \leq 4$. | First we let $0 < a < b$ and seek the $x$ that minimizes \[ f(x) = \left(\frac{x}{a} - 1\right)^2 + \left(\frac{b}{x} - 1\right)^2 \] on $a \leq x \leq b$. Let $x/a = z$ and $b/a = c$. Then \[ f(x) = g(z) = (z - 1)^2 + \left(\frac{c}{z} - 1\right)^2. \] Now $g'(z) = 0$ implies \[ z^4 - z^3 + cz - c^2 = (z^2 - c)(z^2 - z + c) = 0; \] the only positive solution is $z = \sqrt{c}$. Since $0 < a < b$, $c > 1$, $\sqrt{c} > 1$, and \[ g(1) = g(c) = (c - 1)^2 + (\sqrt{c} - 1)^2(\sqrt{c} + 1)^2 > 2(\sqrt{c} - 1). \] Hence the minimum of $g(z)$ on $1 \leq z \leq c$ occurs at $z = \sqrt{c}$. It follows that the minimum for $f(x)$ on $a \leq x \leq b$ occurs at $x = a\sqrt{b/a} = \sqrt{ab}$. Then the minimum for the given function of $r, s, t$ occurs with $r = \sqrt{5}, t = \sqrt{4s}, r = \sqrt{2}$. These imply that $r = \sqrt{2}, s = 2, t = 2\sqrt{2}$. Thus the desired minimum value is \[ 4(\sqrt{2} - 1)^2 = \boxed{12 - 8\sqrt{2}}. \] | 0 |
1981 | 1981_B5 | Let $B(n)$ be the number of ones in the base two expression for the positive integer $n$. For example, $B(6) = B(110_2) = 2$ and $B(15) = B(1111_2) = 4$. Determine whether or not \[ \exp \left( \sum_{n=1}^\infty \frac{B(n)}{n(n+1)} \right) \] is a rational number and find its value. Here $\exp(x)$ denotes $e^x$. | If $n$ has $d$ digits in base 2, $2^{d-1} \leq n$ and so \[ B(n) \leq d \leq 1 + \log_2 n. \] This readily implies that $\sum_{n=1}^\infty \frac{B(n)}{n(n+1)}$ converges to a real number $S$. Hence the manipulations below with convergent series are allowable in the two solutions which follow.
Each $n$ is uniquely expressible as $n_0 + 2n_1 + 2^2n_2 + \cdots$ with each $n_i \in \{0,1\}$ (and with $n_i = 0$ for all but a finite set of $i$). Since \[ 1 + 2 + 2^2 + \cdots + 2^{l-1} = 2^l - 1, \] one sees that $n_i = 1$ if and only if $n$ is of the form $k + 2^l + 2^{l+1}j$ with $k \in \{0, 1, \ldots, 2^l - 1\}$ and $j \in \{0, 1, 2, \ldots\}$. Thus
\[ S = \sum_{n=1}^\infty \frac{B(n)}{n(n+1)} = \sum_{l=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^{2^l - 1} \frac{1}{(k + 2^l + 2^{l+1}j)(1 + k + 2^l + 2^{l+1}j)}. \]
Using $1/s(s+1) = 1/s - 1/(s+1)$, the innermost sum telescopes and
\[ S = \sum_{l=0}^\infty \sum_{j=0}^\infty \left[ \frac{1}{2^l(1 + 2j)} - \frac{1}{2^l(2 + 2j)} \right] = \sum_{l=0}^\infty \frac{1}{2^l} \sum_{j=0}^\infty \frac{(-1)^j}{j}. \]
Since it is well known that $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$, \[ S = \left( \sum_{l=0}^\infty \frac{1}{2^l} \right) \ln 2 = 2 \ln 2 = \ln 4, \] and $e^S$ is the rational number \boxed{4}. | numerical | putnam (modified boxing) | Number Theory Analysis | Let $B(n)$ be the number of ones in the base two expression for the positive integer $n$. For example, $B(6) = B(110_2) = 2$ and $B(15) = B(1111_2) = 4$. Determine whether or not \[ \exp \left( \sum_{n=1}^\infty \frac{B(n)}{n(n+1)} \right) \] is a rational number. Here $\exp(x)$ denotes $e^x$. | If $n$ has $d$ digits in base 2, $2^{d-1} \leq n$ and so \[ B(n) \leq d \leq 1 + \log_2 n. \] This readily implies that $\sum_{n=1}^\infty \frac{B(n)}{n(n+1)}$ converges to a real number $S$. Hence the manipulations below with convergent series are allowable in the two solutions which follow.
Each $n$ is uniquely expressible as $n_0 + 2n_1 + 2^2n_2 + \cdots$ with each $n_i \in \{0,1\}$ (and with $n_i = 0$ for all but a finite set of $i$). Since \[ 1 + 2 + 2^2 + \cdots + 2^{l-1} = 2^l - 1, \] one sees that $n_i = 1$ if and only if $n$ is of the form $k + 2^l + 2^{l+1}j$ with $k \in \{0, 1, \ldots, 2^l - 1\}$ and $j \in \{0, 1, 2, \ldots\}$. Thus
\[ S = \sum_{n=1}^\infty \frac{B(n)}{n(n+1)} = \sum_{l=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^{2^l - 1} \frac{1}{(k + 2^l + 2^{l+1}j)(1 + k + 2^l + 2^{l+1}j)}. \]
Using $1/s(s+1) = 1/s - 1/(s+1)$, the innermost sum telescopes and
\[ S = \sum_{l=0}^\infty \sum_{j=0}^\infty \left[ \frac{1}{2^l(1 + 2j)} - \frac{1}{2^l(2 + 2j)} \right] = \sum_{l=0}^\infty \frac{1}{2^l} \sum_{j=0}^\infty \frac{(-1)^j}{j}. \]
Since it is well known that $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$, \[ S = \left( \sum_{l=0}^\infty \frac{1}{2^l} \right) \ln 2 = 2 \ln 2 = \ln 4, \] and $e^S$ is the rational number 4. | 0 |
1981 | 1981_B6 | Let $C$ be a fixed unit circle in the Cartesian plane. For any convex polygon $P$ each of whose sides is tangent to $C$, let $N(P, h, k)$ be the number of points common to $P$ and the unit circle with center at $(h, k)$. Let $H(P)$ be the region of all points $(x, y)$ for which $N(P, x, y) \geq 1$ and $F(P)$ be the area of $H(P)$. Find the smallest number $u$ with \[ \frac{1}{F(P)} \int \int_{H(P)} N(P, x, y) \ dx \ dy < u \] for all polygons $P$, where the double integral is taken over $H(P)$. | Let $L' = L(P)$ be the perimeter of $P$. One sees that $H(P)$ consists of the region bounded by $P$, the regions bounded by rectangles whose bases are the sides of $P$ and whose altitudes equal 1, and sectors of unit circles which can be put together to form one unit circle. Hence \[ F(P) = \frac{L}{2} + L + \pi = \pi + \frac{3L}{2}. \] If $A$ and $B$ are two consecutive vertices of $P$, the contribution of side $AB$ to the double integral \[ I \] is double the area of the region (of the figure) bounded by the unit semicircles with centers at $A$ and $B$ and segments $CD$ and $EF$ such that $ABCD$ and $ABEF$ are rectangles and $|AD| = |AF| = 1$. One doubles this area because there is a symmetric region bounded by $CD$, $EF$, and the other halves of the unit circles centered at $A$ and $B$. The overlap of the two regions counts twice. By Cavalieri's slicing principle, this contribution of side $AB$ to $I$ is 4 times the length of $AB$. Hence \[ I = 4L \] and \[ \frac{I}{F(P)} = \frac{4L}{\pi + 3L/2} = \frac{8}{3 + (2\pi/L)}. \] One can make $L$ arbitrarily large (e.g., by letting $P$ be a triangle with two angles arbitrarily close to right angles). Hence the desired least upper bound is $\boxed{8/3}$. | numerical | putnam | Geometry Analysis | Let $C$ be a fixed unit circle in the Cartesian plane. For any convex polygon $P$ each of whose sides is tangent to $C$, let $N(P, h, k)$ be the number of points common to $P$ and the unit circle with center at $(h, k)$. Let $H(P)$ be the region of all points $(x, y)$ for which $N(P, x, y) \geq 1$ and $F(P)$ be the area of $H(P)$. Find the smallest number $u$ with \[ \frac{1}{F(P)} \int \int_{H(P)} N(P, x, y) \ dx \ dy < u \] for all polygons $P$, where the double integral is taken over $H(P)$. | Let $L' = L(P)$ be the perimeter of $P$. One sees that $H(P)$ consists of the region bounded by $P$, the regions bounded by rectangles whose bases are the sides of $P$ and whose altitudes equal 1, and sectors of unit circles which can be put together to form one unit circle. Hence \[ F(P) = \frac{L}{2} + L + \pi = \pi + \frac{3L}{2}. \] If $A$ and $B$ are two consecutive vertices of $P$, the contribution of side $AB$ to the double integral \[ I \] is double the area of the region (of the figure) bounded by the unit semicircles with centers at $A$ and $B$ and segments $CD$ and $EF$ such that $ABCD$ and $ABEF$ are rectangles and $|AD| = |AF| = 1$. One doubles this area because there is a symmetric region bounded by $CD$, $EF$, and the other halves of the unit circles centered at $A$ and $B$. The overlap of the two regions counts twice. By Cavalieri's slicing principle, this contribution of side $AB$ to $I$ is 4 times the length of $AB$. Hence \[ I = 4L \] and \[ \frac{I}{F(P)} = \frac{4L}{\pi + 3L/2} = \frac{8}{3 + (2\pi/L)}. \] One can make $L$ arbitrarily large (e.g., by letting $P$ be a triangle with two angles arbitrarily close to right angles). Hence the desired least upper bound is $\boxed{8/3}$. | 0 |
1982 | 1982_A1 | Let $V$ be the region in the cartesian plane consisting of all points $(x, y)$ satisfying the simultaneous conditions \[ |x| \leq y \leq |x| + 3 \text{ and } y \leq 4. \] Find the centroid $(\bar{x}, \bar{y})$ of $V$. | Let $T$ consist of the points inside or on the triangle with vertices at $(0,3), (-1,4), (1,4)$ and let $U$ be the set of points inside or on the triangle with vertices at $(0,0), (-4,4), (4,4)$. Then $T$ and $V$ overlap only on boundary points and their union is $U$. The centroids of $T$ and $U$ are $(0, 11/3)$ and $(0, 8/3)$, respectively. The areas of $T$, $V$, and $U$ are $1$, $15$, and $16$, respectively. Using weighted averages with the areas as weights, one has\[ 1 \cdot 0 + 15\bar{x} = 16 \cdot 0, \]\[ 1 \cdot \frac{11}{3} + 15\bar{y} = 16 \cdot \frac{8}{3}. \] It follows that $\bar{x} = 0, \bar{y} = 13/5$ giving us the centroid at \boxed{(0, 13/5)}. | numerical | putnam | Geometry | Let $V$ be the region in the cartesian plane consisting of all points $(x, y)$ satisfying the simultaneous conditions \[ |x| \leq y \leq |x| + 3 \text{ and } y \leq 4. \] Find the centroid $(\bar{x}, \bar{y})$ of $V$. | Let $T$ consist of the points inside or on the triangle with vertices at $(0,3), (-1,4), (1,4)$ and let $U$ be the set of points inside or on the triangle with vertices at $(0,0), (-4,4), (4,4)$. Then $T$ and $V$ overlap only on boundary points and their union is $U$. The centroids of $T$ and $U$ are $(0, 11/3)$ and $(0, 8/3)$, respectively. The areas of $T$, $V$, and $U$ are $1$, $15$, and $16$, respectively. Using weighted averages with the areas as weights, one has\[ 1 \cdot 0 + 15\bar{x} = 16 \cdot 0, \]\[ 1 \cdot \frac{11}{3} + 15\bar{y} = 16 \cdot \frac{8}{3}. \] It follows that $\boxed{\bar{x} = 0, \bar{y} = 13/5}.$ | 0 |
1982 | 1982_A3 | Evaluate \[ \int_0^\infty \frac{\arctan(\pi x) - \arctan(x)}{x} \, dx. \] | \[ \int_0^\infty \frac{\arctan(\pi x) - \arctan(x)}{x} \, dx = \int_0^\infty \frac{1}{x} \int_1^\pi \arctan(ux) \, du \, dx \\ = \int_1^\pi \int_0^\infty \frac{1}{1 + (xu)^2} \, dx \, du = \int_1^\pi \frac{\pi}{2u} \, du \\ = \int_1^\pi \frac{1}{u} \, du \cdot \frac{\pi}{2} = \boxed{\frac{\pi}{2} \ln \pi}. \] | numerical | putnam | Calculus | Evaluate \[ \int_0^\infty \frac{\arctan(\pi x) - \arctan(x)}{x} \, dx. \] | \[ \int_0^\infty \frac{\arctan(\pi x) - \arctan(x)}{x} \, dx = \int_0^\infty \frac{1}{x} \int_1^\pi \arctan(ux) \, du \, dx \\ = \int_1^\pi \int_0^\infty \frac{1}{1 + (xu)^2} \, dx \, du = \int_1^\pi \frac{\pi}{2u} \, du \\ = \int_1^\pi \frac{1}{u} \, du \cdot \frac{\pi}{2} = \boxed{\frac{\pi}{2} \ln \pi}. \] | 0 |
1982 | 1982_A5 | Let $a, b, c,$ and $d$ be positive integers and
\[r = 1 - \frac{a}{b} - \frac{c}{d}.\]
Given that $a + c \leq 1982$ and $r > 0$, find the lowerbound of r.\] | We are given that
\[r = 1 - \frac{a}{b} - \frac{c}{d} = \frac{bd - ad - bc}{bd} > 0.\]
Thus $bd - ad - bc$ is a positive integer and so $r \geq 1/bd$. We may assume without loss of generality that $b \leq d$. If $b \leq d \leq 1983$, $r \geq 1983^{-2} > 1983^{-3}$. Since $a + c \leq 1982$, if $1983 \leq b \leq d$, one has
\[r \geq 1 - \frac{a}{1983} - \frac{c}{1983} > 1 - \frac{1982}{1983} = \frac{1}{1983} > \frac{1}{1983^3}.\]
The remaining case is that with $b < 1983 < d$. Then the $d$ that minimizes $r$ for fixed $a, b, c$ is
\[1 + \frac{bc}{\lfloor b - a \rfloor},\]
where $\lfloor x \rfloor$ is the greatest integer in $x$. This $d$ is at most $1983b$ since $b - a \geq 1$ and $c < 1982$ and thus
\[r \geq \frac{1}{bd} \geq \frac{1}{1983b^2} > \boxed{\frac{1}{1983^3}}.\]
Hence we have the desired inequality in all cases. | numerical | putnam (modified boxing) | Algebra | Let $a, b, c,$ and $d$ be positive integers and
\[r = 1 - \frac{a}{b} - \frac{c}{d}.\]
Given that $a + c \leq 1982$ and $r > 0$, prove that
\[r > \frac{1}{1983^3}.\] | We are given that
\[r = 1 - \frac{a}{b} - \frac{c}{d} = \frac{bd - ad - bc}{bd} > 0.\]
Thus $bd - ad - bc$ is a positive integer and so $r \geq 1/bd$. We may assume without loss of generality that $b \leq d$. If $b \leq d \leq 1983$, $r \geq 1983^{-2} > 1983^{-3}$. Since $a + c \leq 1982$, if $1983 \leq b \leq d$, one has
\[r \geq 1 - \frac{a}{1983} - \frac{c}{1983} > 1 - \frac{1982}{1983} = \frac{1}{1983} > \frac{1}{1983^3}.\]
The remaining case is that with $b < 1983 < d$. Then the $d$ that minimizes $r$ for fixed $a, b, c$ is
\[1 + \frac{bc}{\lfloor b - a \rfloor},\]
where $\lfloor x \rfloor$ is the greatest integer in $x$. This $d$ is at most $1983b$ since $b - a \geq 1$ and $c < 1982$ and thus
\[r \geq \frac{1}{bd} \geq \frac{1}{1983b^2} > \frac{1}{1983^3}.\]
Hence we have the desired inequality in all cases. | 0 |
1982 | 1982_B1 | Let $M$ be the midpoint of side $BC$ of a general $\triangle ABC$. Using the smallest possible $n$, describe a method for cutting $\triangle AMB$ into $n$ triangles which can be reassembled to form a triangle congruent to $\triangle AMC$ and return this value of $n$. | Let $D$ be the midpoint of side $AB$. Cut $\triangle AMB$ along $DM$. Then $\triangle BMD$ can be placed alongside $\triangle ADM$, with side $BD$ atop side $AD$, so as to form a triangle congruent to $\triangle AMC$. Since $\triangle AMB$ need not be congruent to $\triangle AMC$ in a general $\triangle ABC$, there is no method with $n = 1$. The smallest $n$ is \boxed{2}. | numerical | putnam | Geometry | Let $M$ be the midpoint of side $BC$ of a general $\triangle ABC$. Using the smallest possible $n$, describe a method for cutting $\triangle AMB$ into $n$ triangles which can be reassembled to form a triangle congruent to $\triangle AMC$. | The smallest $n$ is \boxed{2}. Let $D$ be the midpoint of side $AB$. Cut $\triangle AMB$ along $DM$. Then $\triangle BMD$ can be placed alongside $\triangle ADM$, with side $BD$ atop side $AD$, so as to form a triangle congruent to $\triangle AMC$. Since $\triangle AMB$ need not be congruent to $\triangle AMC$ in a general $\triangle ABC$, there is no method with $n = 1$. | 0 |
1982 | 1982_B2 | Let $A(x, y)$ denote the number of points $(m, n)$ in the plane with integer coordinates $m$ and $n$ satisfying $m^2 + n^2 \leq x^2 + y^2$. Let $g = \sum_{k=0}^\infty e^{-k^2}$. Express
\[\int_{-x}^x \int_{-x}^x A(x, y) e^{-x^2-y^2} \,dx \,dy\]
as a polynomial in $g$. | Let $r = \sqrt{x^2 + y^2}$, $R(m, n) = \{(x, y) : m^2 + n^2 \leq x^2 + y^2\}$, and
\[ I = \int_{-\infty}^\infty \int_{-\infty}^\infty A(x, y) e^{-x^2-y^2} \,dx \,dy. \]
Let $\Sigma$ and $\Sigma'$ denote sums over all integers $m$ and over all integers $n$, respectively. Then
\[ I = \Sigma \Sigma' \int_{R(m, n)} e^{-x^2-y^2} \,dx \,dy \]
\[ = \Sigma \Sigma' \int_0^{2\pi} \int_{\sqrt{m^2+n^2}}^\infty e^{-r^2} r \,dr \,d\theta \]
\[ = \Sigma \Sigma' \int_0^{2\pi} \left[ -\frac{1}{2} e^{-r^2} \right]_{\sqrt{m^2+n^2}}^\infty \,d\theta \]
\[ = \Sigma \Sigma' \int_0^{2\pi} \frac{1}{2} e^{-m^2-n^2} \,d\theta \]
\[ = \Sigma \Sigma' \pi e^{-m^2-n^2} = \pi \left( \sum e^{-m^2} \right) \left( \sum e^{-n^2} \right) = \boxed{\pi (2g - 1)^2}. \] | algebraic | putnam | Algebra Calculus | Let $A(x, y)$ denote the number of points $(m, n)$ in the plane with integer coordinates $m$ and $n$ satisfying $m^2 + n^2 \leq x^2 + y^2$. Let $g = \sum_{k=0}^\infty e^{-k^2}$. Express
\[\int_{-x}^x \int_{-x}^x A(x, y) e^{-x^2-y^2} \,dx \,dy\]
as a polynomial in $g$. | Let $r = \sqrt{x^2 + y^2}$, $R(m, n) = \{(x, y) : m^2 + n^2 \leq x^2 + y^2\}$, and
\[ I = \int_{-\infty}^\infty \int_{-\infty}^\infty A(x, y) e^{-x^2-y^2} \,dx \,dy. \]
Let $\Sigma$ and $\Sigma'$ denote sums over all integers $m$ and over all integers $n$, respectively. Then
\[ I = \Sigma \Sigma' \int_{R(m, n)} e^{-x^2-y^2} \,dx \,dy \]
\[ = \Sigma \Sigma' \int_0^{2\pi} \int_{\sqrt{m^2+n^2}}^\infty e^{-r^2} r \,dr \,d\theta \]
\[ = \Sigma \Sigma' \int_0^{2\pi} \left[ -\frac{1}{2} e^{-r^2} \right]_{\sqrt{m^2+n^2}}^\infty \,d\theta \]
\[ = \Sigma \Sigma' \int_0^{2\pi} \frac{1}{2} e^{-m^2-n^2} \,d\theta \]
\[ = \Sigma \Sigma' \pi e^{-m^2-n^2} = \pi \left( \sum e^{-m^2} \right) \left( \sum e^{-n^2} \right) = \boxed{\pi (2g - 1)^2}. \] | 0 |
1982 | 1982_B3 | Let $p_n$ be the probability that $c + d$ is a perfect square when the integers $c$ and $d$ are selected independently at random from the set $\{1, 2, \ldots, n\}$. Show that $\lim_{n \to \infty}(p_n\sqrt{n})$ exists and express this limit in the form $r(\sqrt{s} - t)$, where $s$ and $t$ are integers and $r$ is a rational number. | Let $a(n) = \lfloor \sqrt{n} + 1 \rfloor$ and $b(n) = \lfloor \sqrt{2n} \rfloor$, where $\lfloor x \rfloor$ is the greatest integer in $x$. For $t$ in $\{1, 2, \ldots, a(n)\}$, there are $t^2 - 1$ ordered pairs $(c, d)$ with $c$ and $d$ in $X = \{1, 2, \ldots, n\}$ and $c + d = t^2$. For $t$ in $\{1 + a(n), 2 + a(n), \ldots, b(n)\}$, there are $2n + 1 - t^2$ ordered pairs $(c, d)$ with $c$ and $d$ in $X$ and $c + d = t^2$. Hence the total number $F(n)$ of favorable $(c, d)$ is
\[ F(n) = \sum_{t=1}^{a(n)} (t^2 - 1) + \sum_{t=1+a(n)}^{b(n)} (2n + 1 - t^2) \]
\[ = \left( 2\sum_{t=1}^{a(n)} t^2 \right) - \left( \sum_{t=1}^{b(n)} t^2 \right) - a(n) + [b(n) - a(n)](2n + 1) \]
\[ = \frac{2a(n)[1 + a(n)][1 + 2a(n)]}{6} - \frac{b(n)[1 + b(n)][1 + 2b(n)]}{6} \]
\[ - 2(n + 1)a(n) + (2n + 1)b(n). \]
Since $p_n = F(n)/n^2$,
\[ \lim_{n \to \infty} \left(p_n\sqrt{n}\right) = \lim_{n \to \infty} \frac{F(n)}{n^{3/2}} \]
\[ = \frac{2}{6} \cdot 2 \cdot \lim_{n \to \infty} \left( \frac{a(n)}{\sqrt{n}} \right)^3 - \frac{2}{6} \cdot \lim_{n \to \infty} \left( \frac{b(n)}{\sqrt{n}} \right)^3 \]
\[ - 2 \cdot \lim_{n \to \infty} \frac{a(n)}{\sqrt{n}} + 2 \cdot \lim_{n \to \infty} \frac{b(n)}{\sqrt{n}} \]
\[ = \frac{2}{3} - \frac{1}{3}(\sqrt{2})^3 - 2 + 2\sqrt{2} \]
\[ = \boxed{\frac{4}{3}(\sqrt{2} - 1)}. \] | numerical | putnam | Probability Number Theory | Let $p_n$ be the probability that $c + d$ is a perfect square when the integers $c$ and $d$ are selected independently at random from the set $\{1, 2, \ldots, n\}$. Show that $\lim_{n \to \infty}(p_n\sqrt{n})$ exists and express this limit in the form $r(\sqrt{s} - t)$, where $s$ and $t$ are integers and $r$ is a rational number. | Let $a(n) = \lfloor \sqrt{n} + 1 \rfloor$ and $b(n) = \lfloor \sqrt{2n} \rfloor$, where $\lfloor x \rfloor$ is the greatest integer in $x$. For $t$ in $\{1, 2, \ldots, a(n)\}$, there are $t^2 - 1$ ordered pairs $(c, d)$ with $c$ and $d$ in $X = \{1, 2, \ldots, n\}$ and $c + d = t^2$. For $t$ in $\{1 + a(n), 2 + a(n), \ldots, b(n)\}$, there are $2n + 1 - t^2$ ordered pairs $(c, d)$ with $c$ and $d$ in $X$ and $c + d = t^2$. Hence the total number $F(n)$ of favorable $(c, d)$ is
\[ F(n) = \sum_{t=1}^{a(n)} (t^2 - 1) + \sum_{t=1+a(n)}^{b(n)} (2n + 1 - t^2) \]
\[ = \left( 2\sum_{t=1}^{a(n)} t^2 \right) - \left( \sum_{t=1}^{b(n)} t^2 \right) - a(n) + [b(n) - a(n)](2n + 1) \]
\[ = \frac{2a(n)[1 + a(n)][1 + 2a(n)]}{6} - \frac{b(n)[1 + b(n)][1 + 2b(n)]}{6} \]
\[ - 2(n + 1)a(n) + (2n + 1)b(n). \]
Since $p_n = F(n)/n^2$,
\[ \lim_{n \to \infty} \left(p_n\sqrt{n}\right) = \lim_{n \to \infty} \frac{F(n)}{n^{3/2}} \]
\[ = \frac{2}{6} \cdot 2 \cdot \lim_{n \to \infty} \left( \frac{a(n)}{\sqrt{n}} \right)^3 - \frac{2}{6} \cdot \lim_{n \to \infty} \left( \frac{b(n)}{\sqrt{n}} \right)^3 \]
\[ - 2 \cdot \lim_{n \to \infty} \frac{a(n)}{\sqrt{n}} + 2 \cdot \lim_{n \to \infty} \frac{b(n)}{\sqrt{n}} \]
\[ = \frac{2}{3} - \frac{1}{3}(\sqrt{2})^3 - 2 + 2\sqrt{2} \]
\[ = \boxed{\frac{4}{3}(\sqrt{2} - 1)}. \] | 0 |
1983 | 1983_A1 | How many positive integers $n$ are there such that $n$ is an exact divisor of at least one of the numbers $10^{40}, 20^{30}$? | For $d$ and $m$ in $\mathbb{Z}^+ = \{1,2,3,\ldots\}$, let $d|m$ denote that $d$ is an integral divisor of $m$. For $m$ in $\mathbb{Z}^+$, let $\tau(m)$ be the number of $d$ in $\mathbb{Z}^+$ such that $d|m$. The number of $n$ in $\mathbb{Z}^+$ such that $n|a$ or $n|b$ is \[ \tau(a) + \tau(b) - \tau(\gcd(a,b)). \] Also $\tau(p^s q^t) = (s+1)(t+1)$ for $p, q$ distinct primes. Thus the desired count is \[ \tau(2^{40} \cdot 5^{40}) + \tau(2^{60} \cdot 5^{30}) - \tau(2^{40} \cdot 5^{30}) = 41^2 + 61 \cdot 31 - 41 \cdot 31 \] \[ = 1681 + 620 = \boxed{2301}. \] | numerical | putnam | Number Theory | How many positive integers $n$ are there such that $n$ is an exact divisor of at least one of the numbers $10^{40}, 20^{30}$? | For $d$ and $m$ in $\mathbb{Z}^+ = \{1,2,3,\ldots\}$, let $d|m$ denote that $d$ is an integral divisor of $m$. For $m$ in $\mathbb{Z}^+$, let $\tau(m)$ be the number of $d$ in $\mathbb{Z}^+$ such that $d|m$. The number of $n$ in $\mathbb{Z}^+$ such that $n|a$ or $n|b$ is \[ \tau(a) + \tau(b) - \tau(\gcd(a,b)). \] Also $\tau(p^s q^t) = (s+1)(t+1)$ for $p, q$ distinct primes. Thus the desired count is \[ \tau(2^{40} \cdot 5^{40}) + \tau(2^{60} \cdot 5^{30}) - \tau(2^{40} \cdot 5^{30}) = 41^2 + 61 \cdot 31 - 41 \cdot 31 \] \[ = 1681 + 620 = \boxed{2301}. \] | 0 |
1983 | 1983_A2 | The hands of an accurate clock have lengths $3$ and $4$. Find the distance between the tips of the hands when that distance is increasing most rapidly. | Let $OA$ be the long hand and $OB$ be the short hand. We can think of $OA$ as fixed and $OB$ as rotating at constant speed. Let $v$ be the vector giving the velocity of point $B$ under this assumption. The rate of change of the distance between $A$ and $B$ is the component of $v$ in the direction of $AB$. Since $v$ is orthogonal to $OB$ and the magnitude of $v$ is constant, this component is maximal when $\angle OBA$ is a right angle, i.e., when the distance $AB$ is $\sqrt{4^2 - 3^2} = \sqrt{7}$. \n Alternatively, let $x$ be the distance $AB$ and $\theta = \angle AOB$. By the Law of Cosines, \n \[ x^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cos \theta = 25 - 24 \cos \theta. \] \n Since $d\theta/dt$ is constant, we may assume units chosen so that $\theta$ is also time $t$. Now \[ 2x \frac{dx}{d\theta} = 24 \sin \theta, \quad \frac{dx}{d\theta} = \frac{12 \sin \theta}{\sqrt{25 - 24 \cos \theta}}. \] \n Since $dx/d\theta$ is an odd function of $\theta$, $|dx/d\theta|$ is a maximum when $dx/d\theta$ is a maximum or a minimum. Since $dx/d\theta$ is a periodic differentiable function of $\theta$, $d^2x/d\theta^2 = 0$ at the extremes for $dx/d\theta$. For such $\theta$, \n \[ 12 \cos \theta = x \frac{d^2x}{d\theta^2} + \left( \frac{dx}{d\theta} \right)^2 = \left( \frac{dx}{d\theta} \right)^2 = \frac{144 \sin^2 \theta}{x^2}. \] \n Then \[ x^2 = \frac{12 \sin^2 \theta}{\cos \theta} = \frac{12 - 12 \cos^2 \theta}{\cos \theta} = 25 - 24 \cos \theta, \] \n and it follows that \[ 12 \cos^2 \theta - 25 \cos \theta + 12 = 0. \] \n The only allowable solution for $\cos \theta$ is $\cos \theta = 3/4$ and hence $x = \sqrt{25 - 24 \cos \theta} = \sqrt{25 - 18} = \boxed{\sqrt{7}}. \] | numerical | putnam | Geometry Trigonometry Calculus | The hands of an accurate clock have lengths $3$ and $4$. Find the distance between the tips of the hands when that distance is increasing most rapidly. | Let $OA$ be the long hand and $OB$ be the short hand. We can think of $OA$ as fixed and $OB$ as rotating at constant speed. Let $v$ be the vector giving the velocity of point $B$ under this assumption. The rate of change of the distance between $A$ and $B$ is the component of $v$ in the direction of $AB$. Since $v$ is orthogonal to $OB$ and the magnitude of $v$ is constant, this component is maximal when $\angle OBA$ is a right angle, i.e., when the distance $AB$ is $\sqrt{4^2 - 3^2} = \sqrt{7}$. \n Alternatively, let $x$ be the distance $AB$ and $\theta = \angle AOB$. By the Law of Cosines, \n \[ x^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cos \theta = 25 - 24 \cos \theta. \] \n Since $d\theta/dt$ is constant, we may assume units chosen so that $\theta$ is also time $t$. Now \[ 2x \frac{dx}{d\theta} = 24 \sin \theta, \quad \frac{dx}{d\theta} = \frac{12 \sin \theta}{\sqrt{25 - 24 \cos \theta}}. \] \n Since $dx/d\theta$ is an odd function of $\theta$, $|dx/d\theta|$ is a maximum when $dx/d\theta$ is a maximum or a minimum. Since $dx/d\theta$ is a periodic differentiable function of $\theta$, $d^2x/d\theta^2 = 0$ at the extremes for $dx/d\theta$. For such $\theta$, \n \[ 12 \cos \theta = x \frac{d^2x}{d\theta^2} + \left( \frac{dx}{d\theta} \right)^2 = \left( \frac{dx}{d\theta} \right)^2 = \frac{144 \sin^2 \theta}{x^2}. \] \n Then \[ x^2 = \frac{12 \sin^2 \theta}{\cos \theta} = \frac{12 - 12 \cos^2 \theta}{\cos \theta} = 25 - 24 \cos \theta, \] \n and it follows that \[ 12 \cos^2 \theta - 25 \cos \theta + 12 = 0. \] \n The only allowable solution for $\cos \theta$ is $\cos \theta = 3/4$ and hence $x = \sqrt{25 - 24 \cos \theta} = \sqrt{25 - 18} = \boxed{\sqrt{7}}. \] | 0 |
1983 | 1983_A6 | Let \( \exp(t) \) denote \( e^t \) and
\[ F(x) = \frac{x^4}{\exp(x^3)} \int_0^x \int_0^{x-u} \exp(u^3 + v^3) \, dv \, du. \]
Find \( \lim_{x \to \infty} F(x) \) or prove that it does not exist. | Under the change of variables \( s = u - v \) and \( t = u + v \), with the Jacobian \( \partial(u, v)/\partial(s, t) = 1/2 \), \( F(x) \) becomes \( I(x)/E(x) \) where
\[ I(x) = \int_0^x \int_{-x}^x \exp\left(\left(\frac{t+s}{2}\right)^3 + \left(\frac{t-s}{2}\right)^3\right) \, ds \, dt
= \int_0^x \int_{-x}^x \exp\left(\frac{1}{4}t^3 + \frac{3}{4}ts^2\right) \, ds \, dt \]
and
\[ E(x) = 2x^{-4}\exp(x^3). \]
Since \( I(x) \) and \( E(x) \) go to \( +\infty \) as \( x \to +\infty \), one can use L’Hôpital’s Rule and we have
\[ \lim_{x \to \infty} F(x) = \lim_{x \to \infty} (I'/E') \]
where
\[ I' = \int_{-x}^x \exp\left(\frac{1}{4}x^3 + \frac{3}{4}xs^2\right) \, ds = \exp(x^3/4) \int_{-x}^x \exp(3xs^2/4) \, ds \]
and
\[ E' = (6x^{-2} - 8x^{-5})\exp(x^3). \]
In the integral for \( I' \), make the change of variable \( s = w/\sqrt{x} \), \( ds = dw/\sqrt{x} \), to obtain
\[ I' = \frac{\exp(x^3/4)}{\sqrt{x}} \int_{-x\sqrt{x}}^{x\sqrt{x}} \exp(3w^2/4) \, dw. \]
Now
\[ \lim_{x \to \infty} F(x) = \lim_{x \to \infty} \frac{I'}{E'} = \lim_{x \to \infty} \frac{\int_{-x\sqrt{x}}^{x\sqrt{x}} \exp(3w^2/4) \, dw}{(6x^{-3/2} - 8x^{-9/2})\exp(x^3/4)}. \]
We can, and do, use L’Hôpital’s rule again to obtain
\[ \lim_{x \to \infty} F(x) = \lim_{x \to \infty} \frac{2(3/2)x^{-1/2}\exp(3x^2/4)}{(27/2)x^{-1/2}\exp(3x^2/4)} = \boxed{\frac{2}{9}}. \] | numerical | putnam | Calculus Analysis | Let \( \exp(t) \) denote \( e^t \) and
\[ F(x) = \frac{x^4}{\exp(x^3)} \int_0^x \int_0^{x-u} \exp(u^3 + v^3) \, dv \, du. \]
Find \( \lim_{x \to \infty} F(x) \) or prove that it does not exist. | Under the change of variables \( s = u - v \) and \( t = u + v \), with the Jacobian \( \partial(u, v)/\partial(s, t) = 1/2 \), \( F(x) \) becomes \( I(x)/E(x) \) where
\[ I(x) = \int_0^x \int_{-x}^x \exp\left(\left(\frac{t+s}{2}\right)^3 + \left(\frac{t-s}{2}\right)^3\right) \, ds \, dt
= \int_0^x \int_{-x}^x \exp\left(\frac{1}{4}t^3 + \frac{3}{4}ts^2\right) \, ds \, dt \]
and
\[ E(x) = 2x^{-4}\exp(x^3). \]
Since \( I(x) \) and \( E(x) \) go to \( +\infty \) as \( x \to +\infty \), one can use L’Hôpital’s Rule and we have
\[ \lim_{x \to \infty} F(x) = \lim_{x \to \infty} (I'/E') \]
where
\[ I' = \int_{-x}^x \exp\left(\frac{1}{4}x^3 + \frac{3}{4}xs^2\right) \, ds = \exp(x^3/4) \int_{-x}^x \exp(3xs^2/4) \, ds \]
and
\[ E' = (6x^{-2} - 8x^{-5})\exp(x^3). \]
In the integral for \( I' \), make the change of variable \( s = w/\sqrt{x} \), \( ds = dw/\sqrt{x} \), to obtain
\[ I' = \frac{\exp(x^3/4)}{\sqrt{x}} \int_{-x\sqrt{x}}^{x\sqrt{x}} \exp(3w^2/4) \, dw. \]
Now
\[ \lim_{x \to \infty} F(x) = \lim_{x \to \infty} \frac{I'}{E'} = \lim_{x \to \infty} \frac{\int_{-x\sqrt{x}}^{x\sqrt{x}} \exp(3w^2/4) \, dw}{(6x^{-3/2} - 8x^{-9/2})\exp(x^3/4)}. \]
We can, and do, use L’Hôpital’s rule again to obtain
\[ \lim_{x \to \infty} F(x) = \lim_{x \to \infty} \frac{2(3/2)x^{-1/2}\exp(3x^2/4)}{(27/2)x^{-1/2}\exp(3x^2/4)} = \boxed{\frac{2}{9}}. \] | 0 |
1983 | 1983_B1 | Let $v$ be a vertex (corner) of a cube $C$ with edges of length 4. Let $S$ be the largest sphere that can be inscribed in $C$. Let $R$ be the region consisting of all points $p$ between $S$ and $C$ such that $p$ is closer to $v$ than to any other vertex of the cube. Find the volume of $R$. | The diameter of $S$ must be 4 and $S$ must be centered at the center of $C$. The set of points inside $C$ nearer to $v$ than to another vertex $w$ is the part of that half-space, bounded by the perpendicular bisector of the segment $vw$, containing $v$ which lies within $C$. The intersection of these sets is a cube $C'$ bounded by the three facial planes of $C$ through $v$ and the three planes which are perpendicular bisectors of the edges of $C$ at $v$. These last 3 planes are planes of symmetry for $C$ and $S$. Hence $R$ is one of 8 disjoint congruent regions whose union is the set of points between $S$ and $C$, excepting those on the 3 planes of symmetry. Therefore \[ 8 \text{ vol}(R) = \text{vol}(C) - \text{vol}(S) = 4^3 - \frac{4\pi}{3} \cdot 2^3, \] \[ \text{vol}(R) = \boxed{8 - \frac{4\pi}{3}}. \] | numerical | putnam | Geometry | Let $v$ be a vertex (corner) of a cube $C$ with edges of length 4. Let $S$ be the largest sphere that can be inscribed in $C$. Let $R$ be the region consisting of all points $p$ between $S$ and $C$ such that $p$ is closer to $v$ than to any other vertex of the cube. Find the volume of $R$. | The diameter of $S$ must be 4 and $S$ must be centered at the center of $C$. The set of points inside $C$ nearer to $v$ than to another vertex $w$ is the part of that half-space, bounded by the perpendicular bisector of the segment $vw$, containing $v$ which lies within $C$. The intersection of these sets is a cube $C'$ bounded by the three facial planes of $C$ through $v$ and the three planes which are perpendicular bisectors of the edges of $C$ at $v$. These last 3 planes are planes of symmetry for $C$ and $S$. Hence $R$ is one of 8 disjoint congruent regions whose union is the set of points between $S$ and $C$, excepting those on the 3 planes of symmetry. Therefore \[ 8 \text{ vol}(R) = \text{vol}(C) - \text{vol}(S) = 4^3 - \frac{4\pi}{3} \cdot 2^3, \] \[ \text{vol}(R) = \boxed{8 - \frac{4\pi}{3}}. \] | 0 |
1983 | 1983_B3 | Assume that the differential equation
\[ y''' + p(x)y'' + q(x)y' + r(x)y = 0 \]
has solutions \(y_1(x), y_2(x), y_3(x)\) on the whole real line such that
\[ y_1^2(x) + y_2^2(x) + y_3^2(x) = 1 \]
for all real \(x\). Let
\[ f(x) = (y_1'(x))^2 + (y_2'(x))^2 + (y_3'(x))^2. \]
Find constants \(A\) and \(B\) such that \(f(x)\) is a solution to the differential equation
\[ y' + Ap(x)y = Br(x) and return the sum. \] | To satisfy the equation, each \(y_i\) must have at least 3 derivatives. Here \(\Sigma\) will be a sum with \(i\) running over 1, 2, 3. We have \(\Sigma y_i^2 = 1\) and \(\Sigma(y_i')^2 = f\). Differentiating, one has \(\Sigma 2y_i y_i' = 0\) and \(\Sigma y_i y_i'' + \Sigma (y_i')^2 = 0\) so \(\Sigma y_i y_i'' = -f\). Differentiating this gives us \(\Sigma y_i y_i''' + \Sigma y_i'(y_i'') = -f'\). This and \(\Sigma y_i'(y_i'') = f'/2\) leads to \(\Sigma y_i y_i''' = -3f'/2\). Multiplying each term of
\[ y_i''' + py_i'' + qy_i' + ry_i = 0 \]
by \(y_i\) and summing gives us
\[-3f'/2 - pf + q \cdot 0 + r = 0. \]
Thus \(f' + (2/3)pf = (2/3)r\) and so \(A = 2/3 = B) making the final answer \boxed{4/3}. | numerical | putnam (modified boxing) | Differential Equations | Assume that the differential equation
\[ y''' + p(x)y'' + q(x)y' + r(x)y = 0 \]
has solutions \(y_1(x), y_2(x), y_3(x)\) on the whole real line such that
\[ y_1^2(x) + y_2^2(x) + y_3^2(x) = 1 \]
for all real \(x\). Let
\[ f(x) = (y_1'(x))^2 + (y_2'(x))^2 + (y_3'(x))^2. \]
Find constants \(A\) and \(B\) such that \(f(x)\) is a solution to the differential equation
\[ y' + Ap(x)y = Br(x). \] | To satisfy the equation, each \(y_i\) must have at least 3 derivatives. Here \(\Sigma\) will be a sum with \(i\) running over 1, 2, 3. We have \(\Sigma y_i^2 = 1\) and \(\Sigma(y_i')^2 = f\). Differentiating, one has \(\Sigma 2y_i y_i' = 0\) and \(\Sigma y_i y_i'' + \Sigma (y_i')^2 = 0\) so \(\Sigma y_i y_i'' = -f\). Differentiating this gives us \(\Sigma y_i y_i''' + \Sigma y_i'(y_i'') = -f'\). This and \(\Sigma y_i'(y_i'') = f'/2\) leads to \(\Sigma y_i y_i''' = -3f'/2\). Multiplying each term of
\[ y_i''' + py_i'' + qy_i' + ry_i = 0 \]
by \(y_i\) and summing gives us
\[-3f'/2 - pf + q \cdot 0 + r = 0. \]
Thus \(f' + (2/3)pf = (2/3)r\) and so \boxed{A = 2/3 = B}. | 0 |
1983 | 1983_B5 | Let \( \|u\| \) denote the distance from the real number \( u \) to the nearest integer. (For example, \( \|2.8\| = 0.2 = \|3.2\| \).) For positive integers \( n \), let \[ a_n = \frac{1}{n} \int_1^n \left\| \frac{n}{x} \right\| dx. \] Determine \( \lim_{n \to \infty} a_n \). You may assume the identity \[ \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots = \frac{\pi}{2}. \] | By definition of \( a_n \) and \( \| \cdot \| \), \[ a_n = \frac{1}{n} \sum_{k=1}^{n-1} \int_{2n/(2k+1)}^{n/k} \left( \frac{n}{x} - k \right) dx + \int_{n/(k+1)}^{2n/(2k+1)} \left( k + 1 - \frac{n}{x} \right) dx. \] Simplifying, \[ a_n = \sum_{k=1}^{n-1} \left[ \ln \frac{2k+1}{2k} - \frac{1}{2k+1} + \frac{1}{2k+1} - \ln \frac{2k+2}{2k+1} \right]. \] Factoring terms, \[ a_n = \ln \prod_{k=1}^{n-1} \frac{(2k+1)^2}{2k(2k+2)} = \ln \left[ \frac{3}{2} \cdot \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} \right]. \] Using the identity \[ \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdots = \frac{\pi}{2}, \] we have \[ \lim_{n \to \infty} a_n = \boxed{\ln(\frac{4}{\pi})}. \] | numerical | putnam | Calculus Analysis | Let \( \|u\| \) denote the distance from the real number \( u \) to the nearest integer. (For example, \( \|2.8\| = 0.2 = \|3.2\| \).) For positive integers \( n \), let \[ a_n = \frac{1}{n} \int_1^n \left\| \frac{n}{x} \right\| dx. \] Determine \( \lim_{n \to \infty} a_n \). You may assume the identity \[ \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots = \frac{\pi}{2}. \] | By definition of \( a_n \) and \( \| \cdot \| \), \[ a_n = \frac{1}{n} \sum_{k=1}^{n-1} \int_{2n/(2k+1)}^{n/k} \left( \frac{n}{x} - k \right) dx + \int_{n/(k+1)}^{2n/(2k+1)} \left( k + 1 - \frac{n}{x} \right) dx. \] Simplifying, \[ a_n = \sum_{k=1}^{n-1} \left[ \ln \frac{2k+1}{2k} - \frac{1}{2k+1} + \frac{1}{2k+1} - \ln \frac{2k+2}{2k+1} \right]. \] Factoring terms, \[ a_n = \ln \prod_{k=1}^{n-1} \frac{(2k+1)^2}{2k(2k+2)} = \ln \left[ \frac{3}{2} \cdot \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} \right]. \] Using the identity \[ \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdots = \frac{\pi}{2}, \] we have \[ \lim_{n \to \infty} a_n = \boxed{\ln(\frac{4}{\pi})}. \] | 0 |
1984 | 1984_A1 | Let $A$ be a solid $a \times b \times c$ rectangular brick in three dimensions, where $a, b, c > 0$. Let $B$ be the set of all points which are a distance at most one from some point of $A$ (in particular, $B$ contains $A$). Express the volume of $B$ as a polynomial in $a, h$, and $c$. | The set $B$ can be partitioned into the following sets: \begin{itemize} \item[(i)] $A$ itself, of volume $abc$; \item[(ii)] two $a \times b \times 1$ bricks, two $a \times c \times 1$ bricks, and two $b \times c \times 1$ bricks, of total volume $2ab + 2ac + 2bc$; \item[(iii)] four quarter-cylinders of length $a$ and radius 1, four quarter-cylinders of length $b$ and radius 1, and four quarter-cylinders of length $c$ and radius 1, of total volume $(a + b + c)\pi$; \item[(iv)] eight spherical sectors, each consisting of one-eighth of a sphere of radius 1, of total volume $4\pi/3$. \end{itemize} Hence the volume of $B$ is \[ \boxed{abc + 2(ab + ac + bc) + \pi(a + b + c) + \frac{4\pi}{3}}. \] | algebraic | putnam | Geometry | Let $A$ be a solid $a \times b \times c$ rectangular brick in three dimensions, where $a, b, c > 0$. Let $B$ be the set of all points which are a distance at most one from some point of $A$ (in particular, $B$ contains $A$). Express the volume of $B$ as a polynomial in $a, h$, and $c$. | The set $B$ can be partitioned into the following sets: \begin{itemize} \item[(i)] $A$ itself, of volume $abc$; \item[(ii)] two $a \times b \times 1$ bricks, two $a \times c \times 1$ bricks, and two $b \times c \times 1$ bricks, of total volume $2ab + 2ac + 2bc$; \item[(iii)] four quarter-cylinders of length $a$ and radius 1, four quarter-cylinders of length $b$ and radius 1, and four quarter-cylinders of length $c$ and radius 1, of total volume $(a + b + c)\pi$; \item[(iv)] eight spherical sectors, each consisting of one-eighth of a sphere of radius 1, of total volume $4\pi/3$. \end{itemize} Hence the volume of $B$ is \[ \boxed{abc + 2(ab + ac + bc) + \pi(a + b + c) + \frac{4\pi}{3}}. \] | 0 |
1984 | 1984_A2 | Express \( \sum_{k=1}^\infty \frac{6^k}{(3^k - 2^k)(3^{k+1} - 2^{k+1})} \) as a rational number. | Let \( S(n) \) denote the \( n \)-th partial sum of the given series. Then \[ S(n) = \sum_{k=1}^n \left[ \frac{3^k}{3^k - 2^k} - \frac{3^{k+1}}{3^{k+1} - 2^{k+1}} \right] = 3 - \frac{3^{n+1}}{3^{n+1} - 2^{n+1}}, \] and the series converges to \( \lim_{n \to \infty} S(n) = \boxed{2} \). | numerical | putnam | Algebra | Express \( \sum_{k=1}^\infty \frac{6^k}{(3^k - 2^k)(3^{k+1} - 2^{k+1})} \) as a rational number. | Let \( S(n) \) denote the \( n \)-th partial sum of the given series. Then \[ S(n) = \sum_{k=1}^n \left[ \frac{3^k}{3^k - 2^k} - \frac{3^{k+1}}{3^{k+1} - 2^{k+1}} \right] = 3 - \frac{3^{n+1}}{3^{n+1} - 2^{n+1}}, \] and the series converges to \( \lim_{n \to \infty} S(n) = \boxed{2} \). | 0 |
1984 | 1984_A3 | Let $n$ be a positive integer. Let $a,b,x$ be real numbers, with $a \neq b$, and let $M_n$ denote the $2n \times 2n$ matrix whose $(i,j)$ entry $m_{i,j}$ is given by
\[
m_{i,j} = \begin{cases}
x & \text{if } i = j, \\
a & \text{if } i \neq j \text{ and } i + j \text{ is even}, \\
b & \text{if } i \neq j \text{ and } i + j \text{ is odd}.
\end{cases}
\]
Thus, for example, $M_2 = \begin{pmatrix}
x & b & a & b \\
b & x & b & a \\
a & b & x & b \\
b & a & b & x
\end{pmatrix}.$
Express $\lim_{x \to a}\det M_n/(x - a)^{2n-2}$ as a polynomial in $a$, $b$, and $n$, where $\det M_n$ denotes the determinant of $M_n$. | Let $N = M_n\big|_{x=a}$. $N$ has rank 2, so that $0$ is an eigenvalue of multiplicity $2n - 2$. Let $e$ denote the $2n \times 1$ column vector of $1$'s. Notice that $Ne = n(a + b)e$, and therefore $n(a + b)$ is an eigenvalue. The trace of $N$ is $2na$, and therefore the remaining eigenvalue is $2na - n(a + b) = n(a - b)$. [Note: This corresponds to the eigenvector $f$, where $f_{i,1} = (-1)^{i+1}, i = 1, \dots, 2n.$]
The preceding analysis implies that the characteristic equation of $N$ is
\[
\det(N - \lambda I) = \lambda^{2n-2}(\lambda - n(a+b))(\lambda - n(a-b)).
\]
Let $\lambda = a - x$. Then
\[
\det M_n = \det(N - (a-x)I) = (a-x)^{2n-2}(a-x-n(a+b))(a-x-n(a-b)).
\]
It follows that
\[
\lim_{x \to a}\frac{\det M_n}{(x-a)^{2n-2}} = \lim_{\lambda \to a}(a-x-n(a+b))(a-x-n(a-b)) = \boxed{n^2(a^2-b^2)}.
\] | algebraic | putnam | Linear Algebra | Let $n$ be a positive integer. Let $a,b,x$ be real numbers, with $a \neq b$, and let $M_n$ denote the $2n \times 2n$ matrix whose $(i,j)$ entry $m_{i,j}$ is given by
\[
m_{i,j} = \begin{cases}
x & \text{if } i = j, \\
a & \text{if } i \neq j \text{ and } i + j \text{ is even}, \\
b & \text{if } i \neq j \text{ and } i + j \text{ is odd}.
\end{cases}
\]
Thus, for example, $M_2 = \begin{pmatrix}
x & b & a & b \\
b & x & b & a \\
a & b & x & b \\
b & a & b & x
\end{pmatrix}.$
Express $\lim_{x \to a}\det M_n/(x - a)^{2n-2}$ as a polynomial in $a$, $b$, and $n$, where $\det M_n$ denotes the determinant of $M_n$. | Let $N = M_n\big|_{x=a}$. $N$ has rank 2, so that $0$ is an eigenvalue of multiplicity $2n - 2$. Let $e$ denote the $2n \times 1$ column vector of $1$'s. Notice that $Ne = n(a + b)e$, and therefore $n(a + b)$ is an eigenvalue. The trace of $N$ is $2na$, and therefore the remaining eigenvalue is $2na - n(a + b) = n(a - b)$. [Note: This corresponds to the eigenvector $f$, where $f_{i,1} = (-1)^{i+1}, i = 1, \dots, 2n.$]
The preceding analysis implies that the characteristic equation of $N$ is
\[
\det(N - \lambda I) = \lambda^{2n-2}(\lambda - n(a+b))(\lambda - n(a-b)).
\]
Let $\lambda = a - x$. Then
\[
\det M_n = \det(N - (a-x)I) = (a-x)^{2n-2}(a-x-n(a+b))(a-x-n(a-b)).
\]
It follows that
\[
\lim_{x \to a}\frac{\det M_n}{(x-a)^{2n-2}} = \lim_{\lambda \to a}(a-x-n(a+b))(a-x-n(a-b)) = \boxed{n^2(a^2-b^2)}.
\] | 0 |
1984 | 1984_A4 | A convex pentagon $P = ABCDE$, with vertices labeled consecutively, is inscribed in a circle of radius 1. Find the maximum area of $P$ subject to the condition that the chords $AC$ and $BD$ be perpendicular. | Let $\theta = \text{Arc } AB$, $\alpha = \text{Arc } DE$, and $\beta = \text{Arc } EA$. Then $\text{Arc } CD = \pi - \theta$ and $\text{Arc } BC = \pi - \alpha - \beta$.
The area of $P$, in terms of the five triangles from the center of the circle is
\[ \frac{1}{2}\sin\theta + \frac{1}{2}\sin(\pi - \theta) + \frac{1}{2}\sin\alpha + \frac{1}{2}\sin\beta + \frac{1}{2}\sin(\pi - \alpha - \beta). \]
This is maximized when $\theta = \pi/2$ and $\alpha = \beta = \pi/3$. Thus, the maximum area is
\[ \frac{1}{2} \cdot 1 \cdot 1 + \frac{1}{2} \cdot 1 \cdot 1 + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \boxed{1 + \frac{3}{4}\sqrt{3}}. \] | numerical | putnam | Geometry | A convex pentagon $P = ABCDE$, with vertices labeled consecutively, is inscribed in a circle of radius 1. Find the maximum area of $P$ subject to the condition that the chords $AC$ and $BD$ be perpendicular. | Let $\theta = \text{Arc } AB$, $\alpha = \text{Arc } DE$, and $\beta = \text{Arc } EA$. Then $\text{Arc } CD = \pi - \theta$ and $\text{Arc } BC = \pi - \alpha - \beta$.
The area of $P$, in terms of the five triangles from the center of the circle is
\[ \frac{1}{2}\sin\theta + \frac{1}{2}\sin(\pi - \theta) + \frac{1}{2}\sin\alpha + \frac{1}{2}\sin\beta + \frac{1}{2}\sin(\pi - \alpha - \beta). \]
This is maximized when $\theta = \pi/2$ and $\alpha = \beta = \pi/3$. Thus, the maximum area is
\[ \frac{1}{2} \cdot 1 \cdot 1 + \frac{1}{2} \cdot 1 \cdot 1 + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \boxed{1 + \frac{3}{4}\sqrt{3}}. \] | 0 |
1984 | 1984_A5 | Let $R$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x + y + z \leq 1$. Let $w = 1 - x - y - z$. Express the value of the triple integral \[ \iiint_R x^1y^9z^8w^4 \, dx \, dy \, dz \] in the form $a!b!c!d!/n!$, where $a, b, c, d,$ and $n$ are positive integers. | For $t > 0$, let $R_t$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x + y + z \leq t$. Let \[ I(t) = \iiint_{R_t} x^1y^9z^8(t - x - y - z)^4 \, dx \, dy \, dz \] and make the change of variables $x = tu, y = tv, z = tw$. We see that $I(t) = I(1)t^{25}$. Let \[ J = \int_0^\infty I(t)e^{-t} \, dt. \] Then \[ J = \int_0^\infty I(1)t^{25}e^{-t} \, dt = I(1)\Gamma(26) = I(1)25!. \] It is also the case that \[ J = \int_0^\infty \iiint_{R_t} e^{-t}x^1y^9z^8(t - x - y - z)^4 \, dx \, dy \, dz \, dt. \] Let $s = t - x - y - z$. Then \[ J = \int_0^\infty \int_0^\infty \int_0^\infty \int_0^\infty e^{-s}e^{-x}e^{-y}e^{-z}x^1y^9z^8s^4 \, dx \, dy \, dz \, ds. \] The integral we desire is \[ I(1) = J/25! = \boxed{1!9!8!4!/25!}. \] | numerical | putnam | Calculus | Let $R$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x + y + z \leq 1$. Let $w = 1 - x - y - z$. Express the value of the triple integral \[ \iiint_R x^1y^9z^8w^4 \, dx \, dy \, dz \] in the form $a!b!c!d!/n!$, where $a, b, c, d,$ and $n$ are positive integers. | For $t > 0$, let $R_t$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x + y + z \leq t$. Let \[ I(t) = \iiint_{R_t} x^1y^9z^8(t - x - y - z)^4 \, dx \, dy \, dz \] and make the change of variables $x = tu, y = tv, z = tw$. We see that $I(t) = I(1)t^{25}$. Let \[ J = \int_0^\infty I(t)e^{-t} \, dt. \] Then \[ J = \int_0^\infty I(1)t^{25}e^{-t} \, dt = I(1)\Gamma(26) = I(1)25!. \] It is also the case that \[ J = \int_0^\infty \iiint_{R_t} e^{-t}x^1y^9z^8(t - x - y - z)^4 \, dx \, dy \, dz \, dt. \] Let $s = t - x - y - z$. Then \[ J = \int_0^\infty \int_0^\infty \int_0^\infty \int_0^\infty e^{-s}e^{-x}e^{-y}e^{-z}x^1y^9z^8s^4 \, dx \, dy \, dz \, ds. \] The integral we desire is \[ I(1) = J/25! = \boxed{1!9!8!4!/25!}. \] | 0 |
1984 | 1984_A6 | Let $n$ be a positive integer, and let $f(n)$ denote the last nonzero digit in the decimal expansion of $n!$. For instance, $f(5) = 2.$ \newline (a) Show that if $a_1, a_2, \dots, a_k$ are distinct nonnegative integers, then $f(5^{a_1} + 5^{a_2} + \dots + 5^{a_k})$ depends only on the sum $a_1 + a_2 + \dots + a_k$. \newline (b) Assuming part (a), we can define \[ g(s) = f(5^{a_1} + 5^{a_2} + \dots + 5^{a_k}), \] where $s = a_1 + a_2 + \dots + a_k$. Find the least positive integer $p$ for which \[ g(s) = g(s + p), \text{ for all } s \geq 1, \] or else show that no such $p$ exists. | \textbf{(a)} All congruences are modulo 10. \newline \textbf{Lemma.} $f(5n) \equiv 2^r f(n).$ \newline \textit{Proof.} We have \newline \[ (5n)! = 10^r n! \prod_{i=0}^{n-1} (5i+1)(5i+2)(5i+3)(5i+4). \] \newline If $i$ is even, then \[ \frac{1}{2}(5i+1)(5i+2)(5i+3)(5i+4) \equiv \frac{1}{2}(1\cdot2\cdot3\cdot4) \equiv 2. \] \newline If $i$ is odd, then \[ \frac{1}{2}(5i+1)(5i+2)(5i+3)(5i+4) \equiv \frac{1}{2}(6\cdot7\cdot8\cdot9) \equiv 2. \] \newline Thus the entire product above is congruent to $2^n$. From $(*)$ it is clear that the largest power of 10 dividing $(5n)!$ is the same as the largest power of 10 dividing $10^r n!$, and the proof follows. \newline \newline We now show by induction on $5^{a_1} + \dots + 5^{a_k}$ that \[ f(5^{a_1} + \dots + 5^{a_k}) \equiv 2^{a_1 + \dots + a_k} \] (which depends only on $a_1 + \dots + a_k$ as desired). \newline \newline This is true for $5^0 = 1$, since $f(5^0) \equiv 2^0 \equiv 1$. \newline \newline \textbf{Case 1.} All $a_i > 0$. By the lemma and induction, \[ f(5^{a_1} + \dots + 5^{a_k}) \equiv 2^{a_1-1}f(5^{a_2-1} + \dots + 5^{a_k-1}) \equiv 2^k \cdot 2^{(a_1-1) + \dots + (a_k-1)} \equiv 2^{a_1 + \dots + a_k}. \] \newline \newline \textbf{Case 2.} Some $a_i = 0$, say $a_1 = 0$. Now \[ f(1 + 5m) \equiv (1 + 5m)f(5m)!. \] \newline So $f(1+5m) \equiv (1+5m)f(5m).$ Letting $m = 5^{a_2-1} + \dots + 5^{a_k-1},$ the proof follows by induction. \newline \newline \textbf{(b)} The least $p > 0$ for which \[ 2^s + p \equiv 2^0 \text{ for all } s \geq 1 \] is $\boxed{4}.$ | numerical | putnam | Algebra Number Theory | Let $n$ be a positive integer, and let $f(n)$ denote the last nonzero digit in the decimal expansion of $n!$. For instance, $f(5) = 2.$ \newline (a) Show that if $a_1, a_2, \dots, a_k$ are distinct nonnegative integers, then $f(5^{a_1} + 5^{a_2} + \dots + 5^{a_k})$ depends only on the sum $a_1 + a_2 + \dots + a_k$. \newline (b) Assuming part (a), we can define \[ g(s) = f(5^{a_1} + 5^{a_2} + \dots + 5^{a_k}), \] where $s = a_1 + a_2 + \dots + a_k$. Find the least positive integer $p$ for which \[ g(s) = g(s + p), \text{ for all } s \geq 1, \] or else show that no such $p$ exists. | \textbf{(a)} All congruences are modulo 10. \newline \textbf{Lemma.} $f(5n) \equiv 2^r f(n).$ \newline \textit{Proof.} We have \newline \[ (5n)! = 10^r n! \prod_{i=0}^{n-1} (5i+1)(5i+2)(5i+3)(5i+4). \] \newline If $i$ is even, then \[ \frac{1}{2}(5i+1)(5i+2)(5i+3)(5i+4) \equiv \frac{1}{2}(1\cdot2\cdot3\cdot4) \equiv 2. \] \newline If $i$ is odd, then \[ \frac{1}{2}(5i+1)(5i+2)(5i+3)(5i+4) \equiv \frac{1}{2}(6\cdot7\cdot8\cdot9) \equiv 2. \] \newline Thus the entire product above is congruent to $2^n$. From $(*)$ it is clear that the largest power of 10 dividing $(5n)!$ is the same as the largest power of 10 dividing $10^r n!$, and the proof follows. \newline \newline We now show by induction on $5^{a_1} + \dots + 5^{a_k}$ that \[ f(5^{a_1} + \dots + 5^{a_k}) \equiv 2^{a_1 + \dots + a_k} \] (which depends only on $a_1 + \dots + a_k$ as desired). \newline \newline This is true for $5^0 = 1$, since $f(5^0) \equiv 2^0 \equiv 1$. \newline \newline \textbf{Case 1.} All $a_i > 0$. By the lemma and induction, \[ f(5^{a_1} + \dots + 5^{a_k}) \equiv 2^{a_1-1}f(5^{a_2-1} + \dots + 5^{a_k-1}) \equiv 2^k \cdot 2^{(a_1-1) + \dots + (a_k-1)} \equiv 2^{a_1 + \dots + a_k}. \] \newline \newline \textbf{Case 2.} Some $a_i = 0$, say $a_1 = 0$. Now \[ f(1 + 5m) \equiv (1 + 5m)f(5m)!. \] \newline So $f(1+5m) \equiv (1+5m)f(5m).$ Letting $m = 5^{a_2-1} + \dots + 5^{a_k-1},$ the proof follows by induction. \newline \newline \textbf{(b)} The least $p > 0$ for which \[ 2^s + p \equiv 2^0 \text{ for all } s \geq 1 \] is $\boxed{4}.$ | 0 |
1984 | 1984_B1 | Let $n$ be a positive integer, and define \[ f(n) = 1! + 2! + \cdots + n!. \] Find polynomials $P(x)$ and $Q(x)$ such that \[ f(n+2) = P(n)f(n+1) + Q(n)f(n), \] for all $n \geq 1 and return their difference. | We have \[ f(n+2) - f(n+1) = (n+2)! = (n+2)(n+1)! = (n+2)[f(n+1) - f(n)]. \] It follows that we can take $P(x) = x+3$ and $Q(x) = -x-2$ with the final answer being \boxed{2x+5}. | algebraic | putnam (modified boxing) | Algebra | Let $n$ be a positive integer, and define \[ f(n) = 1! + 2! + \cdots + n!. \] Find polynomials $P(x)$ and $Q(x)$ such that \[ f(n+2) = P(n)f(n+1) + Q(n)f(n), \] for all $n \geq 1.$ | We have \[ f(n+2) - f(n+1) = (n+2)! = (n+2)(n+1)! = (n+2)[f(n+1) - f(n)]. \] It follows that we can take \boxed{$P(x) = x+3$ and $Q(x) = -x-2.$} | 0 |
1984 | 1984_B2 | Find the minimum value of \[ (u-v)^2 + \left( \sqrt{2} - u^2 - \frac{9}{v} \right)^2 \] for \( 0 < u < \sqrt{2} \text{ and } v > 0 \). | The problem asks for the minimum distance between the quarter of the circle \( x^2 + y^2 = 2 \) in the open first quadrant and the half of the hyperbola \( xy = 9 \) in that quadrant. Since the tangents to the respective curves at \((1,1)\) and \((3,3)\) separate the curves and are both perpendicular to \(x = y\), the minimum distance is \(\boxed{8}.\) | numerical | putnam | Calculus Geometry | Find the minimum value of \[ (u-v)^2 + \left( \sqrt{2} - u^2 - \frac{9}{v} \right)^2 \] for \( 0 < u < \sqrt{2} \text{ and } v > 0 \). | The problem asks for the minimum distance between the quarter of the circle \( x^2 + y^2 = 2 \) in the open first quadrant and the half of the hyperbola \( xy = 9 \) in that quadrant. Since the tangents to the respective curves at \((1,1)\) and \((3,3)\) separate the curves and are both perpendicular to \(x = y\), the minimum distance is \(\boxed{8}.\) | 0 |
1984 | 1984_B5 | For each nonnegative integer $k$, let $d(k)$ denote the number of $1$'s in the binary expansion of $k$ (for example, $d(0) = 0$ and $d(5) = 2$). Let $m$ be a positive integer. Express \[ \sum_{k=0}^{2^m - 1} (-1)^{d(k)} k^m \] in the form $(-1)^m a f^{(m)}(g(m))!$, where $a$ is an integer and $f$ and $g$ are polynomials. | Define \[ D(x) = (1 - x)(1 - x^2)(1 - x^4) \cdots (1 - x^{2^{m-1}}). \] Since binary expansions are unique, each monomial $x^k$ $(0 \leq k \leq 2^m - 1)$ appears exactly once in the expansion of $D(x)$, with coefficient $(-1)^{d(k)}$. That is, \[ D(x) = \sum_{k=0}^{2^m-1} (-1)^{d(k)} x^k. \] Applying the operator \[ \left(x \frac{d}{dx}\right)^m \] to $D(x)$, we obtain \[ \left(x \frac{d}{dx}\right)^m D(x) = \sum_{k=0}^{2^m-1} (-1)^{d(k)} k^m x^k, \] so that \[ \left(x \frac{d}{dx}\right)^m D(x) \big|_{x=1} = \sum_{k=0}^{2^m-1} (-1)^{d(k)} k^m. \] Define $F(x) = D(x+1)$, so that \[ \left(x \frac{d}{dx}\right)^m D(x) \big|_{x=1} = \left[ \left((x+1) \frac{d}{dx}\right)^m F(x) \right]_{x=0}. \] But \[ F(x) = \prod_{\alpha=1}^m \left[ 1 - (x+1)^{2^{\alpha-1}} \right] = \prod_{\alpha=1}^m \left[-2^{2^{\alpha-1}}x + O(x^2)\right], \quad (x \to 0). \] This evaluates to \[ F(x) = (-1)^m 2^{m(m-1)/2} x^m + O(x^{m+1}), \] and by observing that \[ \left((x+1) \frac{d}{dx}\right)^m \cdot x^n = nx^n + nx^{n-1}, \] we see that \[ \left((x+1) \frac{d}{dx}\right)^m F(x) = m!A + O(x). \] Thus, \[ \left[ \left((x+1) \frac{d}{dx}\right)^m F(x) \right]_{x=0} = \boxed{(-1)^m 2^{m(m-1)/2} m!}. \] | algebraic | putnam | Calculus Algebra | For each nonnegative integer $k$, let $d(k)$ denote the number of $1$'s in the binary expansion of $k$ (for example, $d(0) = 0$ and $d(5) = 2$). Let $m$ be a positive integer. Express \[ \sum_{k=0}^{2^m - 1} (-1)^{d(k)} k^m \] in the form $(-1)^m a f^{(m)}(g(m))!$, where $a$ is an integer and $f$ and $g$ are polynomials. | 0 |
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1984 | 1984_B6 | A sequence of convex polygons $\{P_n\}, n \geq 0$, is defined inductively as follows. $P_0$ is an equilateral triangle with sides of length 1. Once $P_n$ has been determined, its sides are trisected; the vertices of $P_{n+1}$ are the \textit{interior} trisection points of the sides of $P_n$. Thus, $P_{n+1}$ is obtained by cutting corners off $P_n$, and $P_n$ has $3 \cdot 2^n$ sides. ($P_1$ is a regular hexagon with sides of length $1/3$.)
Express $\lim_{n \to \infty} \text{Area}(P_n)$ in the form $\sqrt{a}/b$, where $a$ and $b$ are positive integers. | Suppose that $\vec{u}$ and $\vec{v}$ are consecutive edges in $P_n$. Then $\vec{u}/3$, $(\vec{u} + \vec{v})/3$, and $\vec{v}/3$ are consecutive edges in $P_{n+1}$. Further,
\[ \frac{1}{2} \|\vec{u}\| \frac{1}{3} \|\vec{v}\| = \frac{1}{18}\|\vec{u} \times \vec{v}\| \]
is removed at this corner in making $P_{n+1}$. But at the next step, the amount from these three consecutive edges is
\[ \frac{1}{2} \frac{\|\vec{u}\| + \|\vec{v}\|}{9} \frac{\|\vec{u}\| + \|\vec{v}\|}{9} + \frac{1}{2} \frac{\|\vec{u} + \vec{v}\|}{9} \frac{\|\vec{v}\|}{9} = \frac{1}{81}\|\vec{u} \times \vec{v}\|. \]
Thus, the amount removed in the $(k+1)\text{st}$ snip is $2/9$ times the amount removed in the $k\text{th}$.
Note that one-third of the original area is removed at the first step. Thus, the amount removed altogether is
\[ \frac{1}{3}\left[1 + (2/9) + (2/9)^2 + \cdots \right] = \frac{1}{3} \cdot \frac{9}{7} = \frac{3}{7} \]
of the original area. Since the original area is $\sqrt{3}/4$, we have
\[ \lim_{n \to \infty} \text{Area } P_n = \frac{4}{7} \cdot \frac{\sqrt{3}}{4} = \boxed{\frac{\sqrt{3}}{7}}. \] | numerical | putnam | Geometry | A sequence of convex polygons $\{P_n\}, n \geq 0$, is defined inductively as follows. $P_0$ is an equilateral triangle with sides of length 1. Once $P_n$ has been determined, its sides are trisected; the vertices of $P_{n+1}$ are the \textit{interior} trisection points of the sides of $P_n$. Thus, $P_{n+1}$ is obtained by cutting corners off $P_n$, and $P_n$ has $3 \cdot 2^n$ sides. ($P_1$ is a regular hexagon with sides of length $1/3$.)
Express $\lim_{n \to \infty} \text{Area}(P_n)$ in the form $\sqrt{a}/b$, where $a$ and $b$ are positive integers. | Suppose that $\vec{u}$ and $\vec{v}$ are consecutive edges in $P_n$. Then $\vec{u}/3$, $(\vec{u} + \vec{v})/3$, and $\vec{v}/3$ are consecutive edges in $P_{n+1}$. Further,
\[ \frac{1}{2} \|\vec{u}\| \frac{1}{3} \|\vec{v}\| = \frac{1}{18}\|\vec{u} \times \vec{v}\| \]
is removed at this corner in making $P_{n+1}$. But at the next step, the amount from these three consecutive edges is
\[ \frac{1}{2} \frac{\|\vec{u}\| + \|\vec{v}\|}{9} \frac{\|\vec{u}\| + \|\vec{v}\|}{9} + \frac{1}{2} \frac{\|\vec{u} + \vec{v}\|}{9} \frac{\|\vec{v}\|}{9} = \frac{1}{81}\|\vec{u} \times \vec{v}\|. \]
Thus, the amount removed in the $(k+1)\text{st}$ snip is $2/9$ times the amount removed in the $k\text{th}$.
Note that one-third of the original area is removed at the first step. Thus, the amount removed altogether is
\[ \frac{1}{3}\left[1 + (2/9) + (2/9)^2 + \cdots \right] = \frac{1}{3} \cdot \frac{9}{7} = \frac{3}{7} \]
of the original area. Since the original area is $\sqrt{3}/4$, we have
\[ \lim_{n \to \infty} \text{Area } P_n = \frac{4}{7} \cdot \frac{\sqrt{3}}{4} = \boxed{\frac{\sqrt{3}}{7}}. \] | 0 |
1985 | 1985_A2 | Let $T$ be an acute triangle. Inscribe a rectangle $R$ in $T$ with one side along a side of $T$. Then inscribe a rectangle $S$ in the triangle formed by the side of $R$ opposite the side on the boundary of $T$, and the other two sides of $T$, with one side along the side of $R$. For any polygon $X$, let $A(X)$ denote the area of $X$. Find the maximum value, or show that no maximum exists, of $\frac{A(R)+A(S)}{A(T)}$, where $T$ ranges over all triangles and $R,S$ over all rectangles as above. | Label lengths as in the figure. Then \[ \frac{A(R) + A(S)}{A(T)} = \frac{ay + bz}{hx/2}, \] where $h = a + b + c$, the altitude of $T$. By similar triangles, \[ \frac{x}{h} = \frac{y}{b+c} = \frac{z}{c}, \] so \[ \frac{A(R) + A(S)}{A(T)} = \frac{(b+c)x}{h} + \frac{cx}{h} = \frac{2}{h^2}(ab + ac + bc). \] We need to maximize $ab + ac + bc$ subject to $a + b + c = h$. One way to do this is to first fix $a$, so $b + c = h - a$. Then \[ ab + ac + bc = a(h - a) + bc, \] and $bc$ is maximized when $b = c$. We now wish to maximize $2ab + b^2$ subject to $a + 2b = h$. This is a straightforward calculus problem giving $a = b = c = h/3$. Hence the maximum ratio is $\boxed{2/3}$ (independent of $T$). | numerical | putnam | Geometry Calculus | Let $T$ be an acute triangle. Inscribe a rectangle $R$ in $T$ with one side along a side of $T$. Then inscribe a rectangle $S$ in the triangle formed by the side of $R$ opposite the side on the boundary of $T$, and the other two sides of $T$, with one side along the side of $R$. For any polygon $X$, let $A(X)$ denote the area of $X$. Find the maximum value, or show that no maximum exists, of $\frac{A(R)+A(S)}{A(T)}$, where $T$ ranges over all triangles and $R,S$ over all rectangles as above. | 0 |
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1985 | 1985_A3 | Let $d$ be a real number. For each integer $m \geq 0$, define a sequence $\{a_m(j)\}$, $j=0,1,2,\dots$ by the condition \begin{align*} a_m(0) &= d/2^m, \\ a_m(j+1) &= (a_m(j))^2 + 2a_m(j), \qquad j \geq 0. \end{align*} Evaluate $\lim_{n \to \infty} a_n(n)$. | We have $a_n(j+1) + 1 = (a_n(j) + 1)^2$, and hence by induction, \[ a_n(j) + 1 = (a_n(0) + 1)^{2^j}. \] Therefore, \[ \lim_{n \to \infty} a_n(n) = \lim_{n \to \infty} \left( 1 + \frac{d}{2^n} \right)^{2^n} - 1 = \boxed{e^d - 1}. \] | algebraic | putnam | Analysis Calculus | Let $d$ be a real number. For each integer $m \geq 0$, define a sequence $\{a_m(j)\}$, $j=0,1,2,\dots$ by the condition \begin{align*} a_m(0) &= d/2^m, \\ a_m(j+1) &= (a_m(j))^2 + 2a_m(j), \qquad j \geq 0. \end{align*} Evaluate $\lim_{n \to \infty} a_n(n)$. | 0 |
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1985 | 1985_A4 | Define a sequence $\{a_i\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for $i\geq 1$. Which integers between 00 and 99 inclusive occur as the last two digits in the decimal expansion of infinitely many $a_i$? | We wish to consider $a_i \equiv 3^{a_{i-1}} \pmod{100}$. Recall that $a^{\phi(n)} \equiv 1 \pmod{n}$ whenever $a$ is relatively prime to $n$ ($\phi$ is the Euler $\phi$-function). Therefore, since $\phi(100) = 40$, we can find $a_i \pmod{100}$ by knowing $a_{i-1} \pmod{40}$. Similarly, we can know $a_{i-1} \pmod{40}$ by finding $a_{i-2} \pmod{16}$, because $a_{i-1} = 3^{a_{i-2}}$ and $\phi(40) = 16$. Again, $a_{i-2} = 3^{a_{i-3}}$ and $\phi(16) = 8$, and therefore \[ a_{i-3} \equiv 3^{a_{i-4}} \equiv 3^{\text{odd integer}} \equiv 3 \pmod{8}. \] It follows that $a_{i-2} \equiv 3^3 \equiv 11 \pmod{16}$, and from this, \[ a_{i-1} \equiv 3^{11} \equiv 27 \pmod{40}, \] and finally, \[ a_i \equiv 3^{27} \equiv 87 \pmod{100}. \] All of this is valid for all $i \geq 4$. Thus, $\boxed{87}$ is the only integer that occurs as the last two digits in the decimal expansions of infinitely many $a_i$. | numerical | putnam | Number Theory | Define a sequence $\{a_i\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for $i\geq 1$. Which integers between 00 and 99 inclusive occur as the last two digits in the decimal expansion of infinitely many $a_i$? | 0 |
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1985 | 1985_A5 | Let $I_m = \int_0^{2\pi} \cos(x)\cos(2x)\cdots \cos(mx)\,dx$. Find the sum of all integers $m$, $1 \leq m \leq 10$ for which $I_m \neq 0$? | Write \[ I_m = \int_0^{2\pi} \prod_{k=1}^m \left( \frac{e^{ikx} + e^{-ikx}}{2} \right) dx = \sum_{\epsilon_k = \pm 1} \frac{1}{2^m} \int_0^{2\pi} e^{i(\epsilon_1 + 2\epsilon_2 + \cdots + m\epsilon_m)x} dx. \] The integral $\int_0^{2\pi} e^{ilx} dx$ is zero if $l$ is a nonzero integer and is $2\pi$ otherwise. Thus, $I_m \geq 0$, and $I_m \neq 0$ if and only if $0$ can be written in the form $\epsilon_1 + 2\epsilon_2 + \cdots + m\epsilon_m$ for some $\epsilon_1, \epsilon_2, \dots, \epsilon_m \in \{-1, 1\}$. For a sum $\epsilon_1 + 2\epsilon_2 + \cdots + m\epsilon_m$, let $r$ denote the sum of the positive terms and $s$ the sum of the absolute values of the negative terms. Then $r - s = m(m + 1)/2$. A necessary condition for $r = s$ is that $m(m + 1)/2$ be even; that is, that $m \equiv 0$ or $3 \pmod{4}$. Thus, the only candidates satisfying these conditions in $1 \leq m \leq 10$ are $m = 3, 4, 7,$ and $8$. We find that $I_m \neq 0$ for each of these because $1 + 2 - 3 = 0$, $1 - 2 - 3 + 4 = 0$, $1 + 2 - 3 + 4 = 0$, and $(1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) = 0$. Thus the final sum becomes \boxed{22}. | numerical | putnam (modified boxing) | Calculus Number Theory | Let $I_m = \int_0^{2\pi} \cos(x)\cos(2x)\cdots \cos(mx)\,dx$. For which integers $m$, $1 \leq m \leq 10$ is $I_m \neq 0$? | 0 |
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1985 | 1985_A6 | If $p(x)= a_0 + a_1 x + \cdots + a_m x^m$ is a polynomial with real coefficients $a_i$, then set \[ \Gamma(p(x)) = a_0^2 + a_1^2 + \cdots + a_m^2. \] Let $F(x) = 3x^2+7x+2$. Find, with proof, a polynomial $g(x)$ with real coefficients such that \begin{enumerate} \item[(i)] $g(0)=1$, and \item[(ii)] $\Gamma(f(x)^n) = \Gamma(g(x)^n)$ \end{enumerate} for every integer $n \geq 1$. | Note that $\Gamma(p(x)) = \int_0^1 |p(e(\theta))|^2 d\theta$, where $e(\theta) = e^{2\pi i \theta}$. Thus, \[ \Gamma(f(x)^n) = \int_0^1 |f(e(\theta))|^{2n} d\theta = \int_0^1 |3e(\theta) + 1|^{2n}|e(\theta) + 2|^{2n} d\theta. \] But \[ |e(\theta) + 2| = |e(\theta)| \cdot |1 + 2e^{-\theta}| = |1 + 2e^{-\theta}| = \sqrt{1 + 4\cos^2(\theta)}. \] Therefore, \[ \Gamma(f(x)^n) = \int_0^1 |3e(\theta) + 1|^{2n} |1 + 2e(\theta)|^{2n} d\theta = \Gamma(g(x)^n), \] where we have set $g(x) = \boxed{6x^2 + 5x + 1}$. | algebraic | putnam | Algebra Analysis | If $p(x)= a_0 + a_1 x + \cdots + a_m x^m$ is a polynomial with real coefficients $a_i$, then set \[ \Gamma(p(x)) = a_0^2 + a_1^2 + \cdots + a_m^2. \] Let $F(x) = 3x^2+7x+2$. Find, with proof, a polynomial $g(x)$ with real coefficients such that \begin{enumerate} \item[(i)] $g(0)=1$, and \item[(ii)] $\Gamma(f(x)^n) = \Gamma(g(x)^n)$ \end{enumerate} for every integer $n \geq 1$. | 0 |
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1985 | 1985_B1 | Let $k$ be the smallest positive integer for which there exist distinct integers $m_1, m_2, m_3, m_4, m_5$ such that the polynomial \[ p(x) = (x-m_1)(x-m_2)(x-m_3)(x-m_4)(x-m_5) \] has exactly $k$ nonzero coefficients. Find a set of integers $m_1, m_2, m_3, m_4, m_5$ for which this minimum $k$ is achieved and given the sum $k+m_1+m_2+m_3+m_4+m_5$ as the final answer. | Clearly $k > 1$; otherwise $p(x) = x^5$ and $m_1, \ldots, m_5$ are not distinct. Assume $k = 2$, so $p(x) = x^5 + ax^j$ with $0 \leq j \leq 4$. We can't have $j \geq 2$ since then at least two of the $m_i$'s are equal to 0. Hence $p(x) = x^5 + a$ or $p(x) = x(x^4 + a)$ with $a \neq 0$. But $x^5 + a$ and $x^4 + a$ have at most two real zeros. Therefore $k > 3$.\n Set $m_1 = -2$, $m_2 = -1$, $m_3 = 0$, $m_4 = 1$, $m_5 = 2$. Then\n \[ p(x) = x(x^2 - 1)(x^2 - 4) = x^5 - 5x^3 + 4x. \] Hence $k = 3$, and this value of $k$ is achieved for the given $m_i$'s. Thus the final sum of $k+m_1+m_2+m_3+m_4+m_5 = \boxed{3}$. | numerical | putnam (modified boxing) | Algebra Number Theory | Let $k$ be the smallest positive integer for which there exist distinct integers $m_1, m_2, m_3, m_4, m_5$ such that the polynomial \[ p(x) = (x-m_1)(x-m_2)(x-m_3)(x-m_4)(x-m_5) \] has exactly $k$ nonzero coefficients. Find, with proof, a set of integers $m_1, m_2, m_3, m_4, m_5$ for which this minimum $k$ is achieved. | 0 |
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1985 | 1985_B2 | Define polynomials $f_n(x)$ for $n \geq 0$ by $f_0(x)=1$, $f_n(0)=0$ for $n \geq 1$, and \[ \frac{d}{dx} f_{n+1}(x) = (n+1)f_n(x+1) \] for $n \geq 0$. Find the explicit factorization of $f_{100}(1)$ into powers of distinct primes. | An examination of low order cases leads one to conjecture that $f_n(x) = x(x+n)^{n-1}$. Clearly this guess satisfies $f_0(x) = 1$, $f_n(0) = 0$ for $n \geq 1$. Now\n \[ f'_{n+1}(x) = (x + n + 1)^n + nx(x + n + 1)^{n-1} \] \[ = (n+1)(x+1)(x+n+1)^{n-1} = (n+1)f_n(x+1). \]\n Hence $f_n(x) = x(x + n)^{n-1}$ as guessed. Therefore, $f_{100}(1) = \boxed{101^{99}}$. | numerical | putnam | Algebra Number Theory | Define polynomials $f_n(x)$ for $n \geq 0$ by $f_0(x)=1$, $f_n(0)=0$ for $n \geq 1$, and \[ \frac{d}{dx} f_{n+1}(x) = (n+1)f_n(x+1) \] for $n \geq 0$. Find, with proof, the explicit factorization of $f_{100}(1)$ into powers of distinct primes. | 0 |
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1985 | 1985_B4 | Let $C$ be the unit circle $x^2+y^2=1$. A point $p$ is chosen randomly on the circumference $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$ and $y$-axes with diagonal $pq$. What is the probability that no point of $R$ lies outside of $C$? | Let $p = (\cos \theta, \sin \theta)$ and $q = (x, y)$. The other two vertices of $R$ are $(\cos \theta, y)$ and $(x, \sin \theta)$, so no point of $R$ lies outside of $C$ if and only if $\cos^2 \theta + y^2 \leq 1$ and $\sin^2 \theta + x^2 \leq 1$, or equivalently, $|y| \leq |\sin \theta|$ and $|x| \leq |\cos \theta|$. Note that these conditions imply that $(x, y)$ lies inside the circle, so that, for any $\theta$, the probability that $(x, y)$ satisfies these conditions is \[ \frac{2|\sin \theta||2|\cos \theta||}{\pi} = \frac{2}{\pi} |\sin 2\theta|, \] and the overall probability is \[ \frac{1}{2\pi} \int_0^{2\pi} \frac{2}{\pi}|\sin 2\theta| d\theta = \frac{1}{2} \cdot \frac{2}{\pi} \cdot 4 = \boxed{\frac{4}{\pi^2}}. \] | numerical | putnam | Geometry Probability | Let $C$ be the unit circle $x^2+y^2=1$. A point $p$ is chosen randomly on the circumference $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$ and $y$-axes with diagonal $pq$. What is the probability that no point of $R$ lies outside of $C$? | 0 |
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1985 | 1985_B5 | Evaluate $\int_0^\infty t^{-1/2}e^{-1985(t+t^{-1})}\,dt$. You may assume that $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$. | Let $I(x) = \int_0^\infty t^{-1/2}e^{-at-x/t}\, dt$, where $a = 1985$. Then \[ I'(x) = -\int_0^\infty t^{-3/2}e^{-at-x/t}\, dt. \] Make the substitution $u = 1/t$, and the last equation becomes \[ I'(x) = -\int_0^\infty u^{-1/2}e^{-au-xu}\, du. \] Now let $w = \frac{x}{a}u$, and the last equation is \[ I'(x) = -\left(\frac{a}{x}\right)^{1/2} \int_0^\infty w^{-1/2}e^{-xw-aw}\, dw = -\left(\frac{a}{x}\right)^{1/2} I(x). \] Therefore, $\log I(x) = -2(ax)^{1/2} + C$, or equivalently, $I(x) = ke^{-2(ax)^{1/2}}$. Also, \[ k = I(0) = \int_0^\infty t^{-1/2}e^{-at}\, dt = \int_0^\infty 2e^{-at^2}\, dt = \frac{\sqrt{\pi}}{\sqrt{a}}. \] This yields \[ I(a) = \boxed{\frac{\sqrt{\pi}}{\sqrt{1985}}e^{-3970}}. \] (Note: the integral is essentially the K-Bessel function $K_{1/2}(3970)$.) | numerical | putnam | Calculus Analysis | Evaluate $\int_0^\infty t^{-1/2}e^{-1985(t+t^{-1})}\,dt$. You may assume that $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$. | 0 |
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1986 | 1986_A1 | Find, with explanation, the maximum value of $f(x)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36\leq 13x^2$. | The condition that $x^4 + 36 \leq 13x^2$ is equivalent to \[ (x - 3)(x - 2)(x + 2)(x + 3) \leq 0. \] The latter is satisfied if and only if $x$ is in the closed interval $[-3, -2]$ or the closed interval $[2, 3]$. The function $f$ is increasing on these intervals because for such $x$, $f'(x) = 3(x^2 - 1) > 0$. It follows that the maximum value of $f$ over this domain is $\max \{f(-2), f(3)\} = \boxed{18}.$ | numerical | putnam | Calculus Algebra | Find, with explanation, the maximum value of $f(x)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36\leq 13x^2$. | 0 |
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1986 | 1986_A2 | What is the units (i.e., rightmost) digit of \[ \left\lfloor \frac{10^{20000}}{10^{100}+3}\right\rfloor ? \] %Here $\lfloor x \rfloor$ is the greatest integer less than or equal to %$x$. | The greatest integer is \[ I = \frac{10^{20000} - 3^{200}}{10^{100} + 3} \] since the remainder is \[ \frac{3^{200}}{10^{100} + 3} < 1. \] We have \[ I \equiv \frac{-3^{200}}{3} \pmod{10} \equiv -3^{199} \pmod{10} \equiv -3^3(3^4)^{49} \pmod{10} \equiv -27 \pmod{10} \equiv 3 \pmod{10}. \] The last digit is \boxed{3}. | numerical | putnam | Number Theory | What is the units (i.e., rightmost) digit of \[ \left\lfloor \frac{10^{20000}}{10^{100}+3}\right\rfloor ? \] %Here $\lfloor x \rfloor$ is the greatest integer less than or equal to %$x$. | 0 |
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1986 | 1986_A3 | Evaluate $\sum_{n=0}^\infty \mathrm{Arccot}(n^2+n+1)$, where $\mathrm{Arccot}\,t$ for $t \geq 0$ denotes the number $\theta$ in the interval $0 < \theta \leq \pi/2$ with $\cot \theta = t$. | Using \[ \cot(\alpha - \beta) = \frac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}, \] one sees that \[ \mathrm{Arccot}(1 + n + n^2) = \mathrm{Arccot} \, n - \mathrm{Arccot} (n + 1). \] Then the series telescopes to \[ \lim_{n \to \infty} (\mathrm{Arccot} \, 0 - \mathrm{Arccot} (n + 1)) = \boxed{\frac{\pi}{2}}. \] | numerical | putnam | Calculus Trigonometry | Evaluate $\sum_{n=0}^\infty \mathrm{Arccot}(n^2+n+1)$, where $\mathrm{Arccot}\,t$ for $t \geq 0$ denotes the number $\theta$ in the interval $0 < \theta \leq \pi/2$ with $\cot \theta = t$. | 0 |
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1986 | 1986_A4 | A \emph{transversal} of an $n\times n$ matrix $A$ consists of $n$ entries of $A$, no two in the same row or column. Let $f(n)$ be the number of $n \times n$ matrices $A$ satisfying the following two conditions: \begin{enumerate} \item[(a)] Each entry $\alpha_{i,j}$ of $A$ is in the set $\{-1,0,1\}$. \item[(b)] The sum of the $n$ entries of a transversal is the same for all transversals of $A$. \end{enumerate} An example of such a matrix $A$ is \[ A = \left( \begin{array}{ccc} -1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{array} \right). \] Determine with proof a formula for $f(n)$ of the form \[ f(n) = a_1 b_1^n + a_2 b_2^n + a_3 b_3^n + a_4, \] where the $a_i$'s and $b_i$'s are rational numbers. | We first prove:\n\textbf{Lemma.} If an $n \times n$ matrix $(\alpha_{ij})$ satisfies $(b)$, then there are unique numbers $c_1 = 0, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ such that $\alpha_{ij} = c_i + d_j$. Conversely, any such choice of $c_i$'s and $d_j$'s yields a unique matrix $(\alpha_{ij})$ satisfying $(b)$.\n\textbf{Proof.} If $\alpha_{ij} = c_i + d_j$ then any transversal of $A$ sums to \n\[\sum_{i=1}^n c_i + \sum_{j=1}^n d_j,\] so $(b)$ is satisfied.\nConversely, suppose $(\alpha_{ij})$ satisfies $(b)$. Define $d_j = \alpha_{1j} - \alpha_{11}$ and $c_i = \alpha_{i1} - d_1 = \alpha_{i1} - \alpha_{11}$ (so $c_1 = 0$). Since $(b)$ is satisfied, we have $\alpha_{ij} + \alpha_{11} = \alpha_{i1} + \alpha_{1j}$ so $\alpha_{ij} = \alpha_{i1} - \alpha_{11} + \alpha_{1j} = c_i + d_j$, as desired.\nThe $c_i$'s and $d_j$'s are unique since $c_1 = 0$ and $\alpha_{ij} = c_i + d_j$ forces $d_j = \alpha_{1j}$, and then $\alpha_{i1} = c_i + d_1$ forces $c_i = \alpha_{i1} - d_1$. This proves the lemma.\nThus $f(n)$ is equal to the number of $2n$-tuples $(c_1 = 0, c_2, \ldots, c_n, d_1, \ldots, d_n)$ for which $c_i + d_j = 0, \pm 1$. We break the possibilities into the following eight cases.\n\[\begin{array}{cccc}\text{distinct $c_i$'s} & \text{possible $d_j$'s} & \text{number of $c_i$'s} & \text{number of $d_j$'s} & \text{product} \\ \hline 0 & 0, -1, 1 & 1 & 3^n & 3^n \\ 0, -2 & 1 & 2^{n-1} - 1 & 1 & 3^{n-1} - 2^n + 1 \\ 0, -2, -1 & 1 & 3^{n-1} - 2^n + 1 & 1 & 3^{n-1} - 2^n + 1 \\ 0, 2 & -1 & 2^{n-1} - 1 & 1 & 2^{n-1} - 1 \\ 0, 1, 2 & -1 & 3^{n-1} - 2^n + 1 & 1 & 3^{n-1} - 2^n + 1 \\ 0, 1 & 0, -1 & 2^{n-1} - 1 & 2^n & 2^{n-1} - 2^n \\ 0, -1 & 0, 1 & 2^{n-1} - 1 & 2^n & 2^{n-1} - 2^n \\ 0, -1, 1 & 0 & 3^{n-1} - 2^n + 1 & 1 & 3^{n-1} - 2^n + 1 \\ \end{array}\] Summing the last column gives \[f(n) = \boxed{4^n + 2 \cdot 3^n - 4 \cdot 2^n + 1}.\] | algebraic | putnam | Combinatorics Linear Algebra | A \emph{transversal} of an $n\times n$ matrix $A$ consists of $n$ entries of $A$, no two in the same row or column. Let $f(n)$ be the number of $n \times n$ matrices $A$ satisfying the following two conditions: \begin{enumerate} \item[(a)] Each entry $\alpha_{i,j}$ of $A$ is in the set $\{-1,0,1\}$. \item[(b)] The sum of the $n$ entries of a transversal is the same for all transversals of $A$. \end{enumerate} An example of such a matrix $A$ is \[ A = \left( \begin{array}{ccc} -1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{array} \right). \] Determine with proof a formula for $f(n)$ of the form \[ f(n) = a_1 b_1^n + a_2 b_2^n + a_3 b_3^n + a_4, \] where the $a_i$'s and $b_i$'s are rational numbers. | 0 |
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